ordinary differential equations - Department of Mathematics ...

ORDINARY DIFFERENTIAL EQUATIONS
GABRIEL NAGY
Mathematics Department,
Michigan State University,
East Lansing, MI, 48824.
JULY 2, 2013
Summary. These notes try to be an introduction to ordinary differential equations. We
describe a collection of methods and techniques used to find solutions to several types
of differential equations. These equations include first order scalar equations, second
order linear equations, and systems of linear equations. There is one chapter dedicated
to power series methods to find solutions to variable coefficients second order linear
equations. There is another chapter dedicated to Laplace transform methods to find
solutions to constant coefficients equations with generalized source functions. The last
chapter provides a brief introduction to boundary value problems and the Fourier transform.
We end the chapter showing how to use these ideas to solve an initial boundary
value problem for a particular partial differential equation, the heat equation.
Date: July 2, 2013, gnagy@math.msu.edu.
1

G. NAGY – ODE July 2, 2013 I
Table of Contents
Chapter 1. First order equations 1
1.1. Linear constant coefficients equations 1
1.1.1. Overview of differential equations 1
1.1.2. Linear equations 2
1.1.3. Linear equations with constant coefficients 3
1.1.4. The Initial Value Problem 6
1.1.5. Exercises 7
1.2. Linear variable coefficients equations 8
1.2.1. Linear equations with variable coefficients 8
1.2.2. The Initial Value Problem 10
1.2.3. The Bernoulli equation 13
1.2.4. Exercises 16
1.3. Separable equations 17
1.3.1. Separable equations 17
1.3.2. Euler Homogeneous equations 22
1.3.3. Exercises 25
1.4. Exact equations 26
1.4.1. Exact differential equations 26
1.4.2. The Poincaré Lemma 27
1.4.3. The integrating factor method 30
1.4.4. Exercises 34
1.5. Applications 35
1.5.1. Radioactive decay 35
1.5.2. Newton’s cooling law 36
1.5.3. Salt in a water tank 36
1.5.4. The Main equations 37
1.5.5. Predictions in particular cases 38
1.5.6. Exercises 42
1.6. Non-linear equations 43
1.6.1. On solutions of linear differential equations 43
1.6.2. On solutions of non-linear differential equations 44
1.6.3. Direction Fields 47
1.6.4. Exercises 50
Chapter 2. Second order linear equations 51
2.1. Variable coefficients 51
2.1.1. Main definitions and properties 51
2.1.2. Wronskian and linear dependence 52
2.1.3. Fundamental and general solutions 54
2.1.4. The Abel Theorem 56
2.1.5. Special Second order nonlinear equations 57
2.1.6. Exercises 60
2.2. Constant coefficients 61
2.2.1. Solution formulas for constant coefficients 61
2.2.2. Different real-valued roots 63
2.2.3. Exercises 65
2.3. Complex roots 66

II G. NAGY – ODE july 2, 2013
2.3.1. Real-valued fundamental solutions 66
2.3.2. The RLC electric circuit 68
2.3.3. Exercises 70
2.4. Repeated roots 71
2.4.1. Fundamental solutions for repeated roots 71
2.4.2. Reduction of order method 73
2.4.3. Exercises 75
2.5. Undetermined coefficients 76
2.5.1. Operator notation 76
2.5.2. The undetermined coefficients method 77
2.5.3. Exercises 80
2.6. Variation of parameters 81
2.6.1. The variation of parameters method 81
2.6.2. Exercises 85
Chapter 3. Power series solutions 86
3.1. Regular points 86
3.1.1. Review of power series 86
3.1.2. Equations with regular points 90
3.1.3. Exercises 96
3.2. The Euler equation 97
3.2.1. The main result 97
3.2.2. The complex-valued roots 99
3.2.3. Exercises 101
3.3. Regular-singular points 102
3.3.1. Finding regular-singular points 102
3.3.2. Solving equations with regular-singular points 104
3.3.3. Exercises 108
Chapter 4. The Laplace Transform 109
4.1. Definition and properties 109
4.1.1. Review of improper integrals 109
4.1.2. Definition of the Laplace Transform 110
4.1.3. Properties of the Laplace Transform 113
4.1.4. Exercises 115
4.2. The initial value problem 116
4.2.1. Summary of the Laplace Transform method 116
4.2.2. Exercises 121
4.3. Discontinuous sources 122
4.3.1. The Step function 122
4.3.2. The Laplace Transform of Step functions 123
4.3.3. The Laplace Transform of translations 124
4.3.4. Differential equations with discontinuous sources 127
4.3.5. Exercises 132
4.4. Generalized sources 133
4.4.1. The Dirac delta generalized function 133
4.4.2. The Laplace Transform of the Dirac delta 135
4.4.3. The fundamental solution 136
4.4.4. Exercises 139

G. NAGY – ODE July 2, 2013 III
4.5. Convolution solutions 140
4.5.1. The convolution of two functions 140
4.5.2. The Laplace Transform of a convolution 141
4.5.3. Impulse response solutions 143
4.5.4. The solution decomposition Theorem 144
4.5.5. Exercises 146
Chapter 5. Systems of differential equations 147
5.1. Introduction 147
5.1.1. Systems of differential equations 147
5.1.2. Reduction to first order systems 150
5.1.3. The Picard-Lindelöf Theorem 151
5.1.4. Exercises 153
5.2. Systems of algebraic equations 154
5.2.1. Linear algebraic systems 154
5.2.2. Gauss elimination operations 158
5.2.3. Linearly dependence 161
5.2.4. Exercises 162
5.3. Matrix algebra 163
5.3.1. A matrix is a function 163
5.3.2. Matrix operations 164
5.3.3. The inverse matrix 168
5.3.4. Determinants 170
5.3.5. Exercises 172
5.4. Linear differential Systems 173
5.4.1. Matrix-vector notation 173
5.4.2. Homogeneous systems 175
5.4.3. Abel’s Theorem for systems 177
5.4.4. Existence of fundamental solutions 179
5.4.5. Exercises 182
5.5. Diagonalizable matrices 183
5.5.1. Eigenvalues and eigenvectors 183
5.5.2. Diagonalizable matrices 189
5.5.3. The exponential of a matrix 193
5.5.4. Exercises 196
5.6. Constant coefficient differential systems 197
5.6.1. Diagonalizable homogeneous systems 197
5.6.2. The Fundamental Solutions 200
5.6.3. Diagonalizable systems with continuous sources 203
5.6.4. Exercises 206
5.7. 2 × 2 Real distinct eigenvalues 207
5.7.1. Phase diagrams 207
5.7.2. Case 0 < λ2 < λ1 208
5.7.3. Case λ 2 < 0 < λ 1 209
5.7.4. Case λ 2 < λ 1 < 0 210
5.7.5. Exercises 212
5.8. 2 × 2 Complex eigenvalues 213
5.8.1. Complex conjugate pairs 213
5.8.2. Qualitative analysis 215
5.8.3. Exercises 216

IV G. NAGY – ODE july 2, 2013
5.9. 2 × 2 Repeated eigenvalues 217
5.9.1. Diagonalizable systems 217
5.9.2. Non-diagonalizable systems 217
5.9.3. Exercises 221
Chapter 6. Boundary value problems 222
6.1. Eigenvalue-eigenfunction problems 222
6.1.1. Comparison: IVP and BVP 222
6.1.2. Eigenvalue-eigenfunction problems 226
6.1.3. Exercises 231
6.2. Overview of Fourier series 232
6.2.1. Origins of Fourier series 232
6.2.2. Fourier series 234
6.2.3. Even and odd functions 238
6.2.4. Exercises 240
6.3. Application: The heat equation 241
6.3.1. The initial-boundary value problem 241
6.3.2. Solution of the initial-boundary value problem 242
6.3.3. Exercises 245
Chapter 7. Appendix 246
7.1. Review exercises 246
7.2. Practice Exams 252
7.3. Answers to exercises 254
References 263

G. NAGY – ODE July 2, 2013 1
Chapter 1. First order equations
A first order differential equations is an equation where the unknown function and only its
first derivative appear in the equation. This Chapter is a collection of methods or recipes to
get solutions to certain types of differential equations, including linear equations, separable
equations, Euler homogeneous equations, and exact equations. The methods we present
here were the first attempts done in the eighteen and nineteen centuries to obtain solutions
to these differential equations.
1.1. Linear constant coefficients equations
1.1.1. Overview of differential equations. A differential equation is an equation where
the unknown is a function and both the function and its derivatives may appear in the
equation. Differential equations are an essential part in the description of nature done by
physics. The central part of many physical theories is a differential equation: Newton’s and
Lagrange equations for classical mechanics, Maxwell’s equations for classical electromagnetism,
Schrödinger’s equation for quantum mechanics, Einstein’s equation for the general
theory of gravitation. We highlight the following examples of differential equations in a
more precise way:
(a) Newton’s second law of motion for a single particle. The unknown is a vector-valued
function x : R → R 3 , where the vector x(t) represents the position of a particle at the
time t. The differential equation is
m d2x (t) = f (t, x(t)),
dt2 where m is a positive constant representing the mass of the particle and f : R×R3 → R3 is the force acting on the particle, which depends on the time t and the position in space
x. This is the well-known “mass times acceleration equals force” law of motion.
(b) The time decay of a radioactive substance. The unknown is a scalar-valued function
u : R → R, where u(t) represents the concentration of the radioactive substance at the
time t. The differential equation is
du
(t) = −k u(t),
dt
where k is a positive constant. The equation says that the higher the material concentration
the faster it decays.
(c) The wave equation, which describes waves propagating in a media. An example is sound,
where pressure waves propagate in the air. The unknown is a scalar-valued function of
two variables u : R × R3 → R, where u(t, x) represents a perturbation in the air density
at the time t and point x = (x, y, z) in space. (We used the same notation for vectors
and points, although they are different type of objects.) The equation is
∂ttu(t, x) = v 2 ∂xxu(t, x) + ∂yyu(t, x) + ∂zzu(t, x) ,
where v is a positive constant describing the wave speed, and we have used the notation
∂ to mean partial derivative.
(d) The heat conduction equation, which describes the variation of temperature in a solid
material. The unknown is a scalar-valued function u : R × R 3 → R, where u(t, x)
represents the temperature at time t and the point x = (x, y, z) in the solid. The
equation is
∂tu(t, x) = k ∂xxu(t, x) + ∂yyu(t, x) + ∂zzu(t, x) ,
where k is a positive constant representing thermal properties of the material.

2 G. NAGY – ODE july 2, 2013
The equations in examples (a) and (b) are called ordinary differential equations (ODE),
since the unknown function depends on a single independent variable, t in these examples.
The equations in examples (c) and (d) are called partial differential equations (PDE), since
the unknown function depends on two or more independent variables and their partial
derivatives appear in the equations.
The order of a differential equation is the highest derivative order that appears in the
equation. Newton’s equation in Example (a) is second order, the time decay equation in
Example (b) is first order, the wave equation in Example (c) and the heat equation in
Example (d) are second order. These last two cases are PDE, and in these cases one usually
gives extra information, saying that the wave equation is second order both in space and
time derivatives but the heat equation is second order in space derivatives and first order
in time derivatives.
1.1.2. Linear equations. We start with a more precise definition of the type of differential
equation we are about to study. When there is no risk of confusion we usually denote with
a prime a derivative of a single variable function. So, the derivative of a function y : R → R
is denoted as
dy
dt (t) = y′ (t).
Definition 1.1.1. Given a function f : R 2 → R, a first order ODE in the unknown
function y : R → R is the equation
y ′ (t) = f(t, y(t)). (1.1.1)
The ODE in Eq. (1.1.1) is called linear iff the function with values f(t, y) is linear on its
second argument y, that is, there exist functions a, b : R → R such that
y ′ (t) = a(t) y(t) + b(t), f(t, y) = a(t) y + b(t). (1.1.2)
Sometimes in the literature there is a different sign convention for Eq. (1.1.2). For
example, Boyce-DiPrima [1] writes it as y ′ = −a y + b. The sign choice in front of function a
is just a convention. Some people like the negative sign, because later on, when they write
the equation as y ′ + a y = b, they get a plus sign on the left-hand side. In any case, we stick
here to the convention y ′ = ay + b.
A linear ODE is called of constant coefficients iff the functions a and b in Eq. (1.1.2) are
both constants. Otherwise, the linear ODE is called of variable coefficients.
Example 1.1.1:
(a) An example of a first order linear ODE is the equation
y ′ (t) = 2 y(t) + 3.
In this case, the right-hand side is given by the function f(t, y) = 2y + 3, where we can
see that a(t) = 2 and b(t) = 3. Since these coefficients do not depend on t, this is a
constant coefficients equation.
(b) Another example of a first order linear ODE is the equation
y ′ (t) = − 2
y(t) + 4t.
t
In this case, the right-hand side is given by the function f(t, y) = −2y/t + 4t, where
a(t) = −2/t and b(t) = 4t. Since the coefficients are non-constant functions of t, this is
a variable coefficients equation.
⊳

G. NAGY – ODE July 2, 2013 3
A function y : D ⊂ R → R is solution of the differential equation in (1.1.1) iff the equation
is satisfied for all values of the independent variable t in the domain D of the function y.
Example 1.1.2: Show that y(t) = e 2t − 3
2 is solution of the ODE y′ (t) = 2 y(t) + 3.
Solution: On the one hand we compute y ′ (t) = 2e2t . On the other hand we compute
2 y(t) + 3 = 2 e 2t − 3
+ 3 = 2e
2
2t .
We conclude that y ′ (t) = 2 y(t) + 3 for all t ∈ R. ⊳
1.1.3. Linear equations with constant coefficients. It is not difficult to obtain all
solutions to first order linear ODE in the case that the equation has constant coefficients.
One multiplies the differential equation by a particularly chosen non-zero function, called the
integrating factor. The integrating factor has the following property: The whole equation
is transformed into a total derivative of a function, called potential function. Integrating
the differential equation is now trivial. The function solution of the differential equation is
simple to obtain inverting the potential function. This whole idea is called the integrating
factor method. In the next Section we generalize this idea to find solutions to variable
coefficients linear ODE, and in Section 1.4 we generalize this idea to certain non-linear
differential equations.
Theorem 1.1.2 (Constant coefficients). If the constants a, b ∈ R satisfy that a = 0,
then linear differential equation
y ′ (t) = a y(t) + b (1.1.3)
has infinitely many solutions and every solution can be labeled by c ∈ R as follows,
y(t) = c e at − b
. (1.1.4)
a
Remark: Theorem 1.1.2 says that Eq. (1.1.3) has infinitely many solutions, one solution for
each value of the constant c, which is not determined by the equation. This is reasonable.
Since the differential equation contains one derivative of the unknown function y, finding
a solution of the differential equation requires to compute an integral. Every indefinite
integration introduces an integration constant. This is the origin of the constant c above.
Proof of Theorem 1.1.2: The integrating factor method is the key idea in the proof of
Theorem 1.1.2. We first write down the differential equation as
y ′ (t) − a y(t) = b,
and then multiply the differential equation by the exponential e −at , where the exponent
is the constant coefficient a in the differential equation multiplied by t. This exponential
function is an integrating factor for the differential equation. The reason to choose this
particular function, e −at , is explained in Lemma 1.1.3, below. The result is
y ′ (t) − a y(t) e −at = b e −at ⇔ e −at y ′ (t) − a e −at y(t) = b e −at .
This exponential is chosen because of the following property,
−a e −at = e −ay ′ .
Introducing this property into the differential equation we get
e −at y ′ (t) + e −at ′ y(t) = b e −at .
Now the product rule for derivatives implies that the left-hand side above is a total derivative,
e −at y(t) ′ = b e −at .

4 G. NAGY – ODE july 2, 2013
At this point we can go one step further, expresing the right-hand side in the differential
equation as b e −at
= − b
a e−at ′
. We obtain
e −at y(t) + b
a e−at ′
= 0 ⇔ y(t) + b
e
a
−at ′
= 0.
Hence, the whole differential equation is a total derivative. The whole differential equation
is the total derivative of the function,
ψ(t, y(t)) = y(t) + b
e
a
−at ,
called potential function. The equation now has the form
dψ
(t, y(t)) = 0.
dt
It is trivial to integrate the differential equation written in terms of the potential function,
ψ(t, y(t)) = c ⇔ y(t) + b
e
a
−at = c.
Simple algebraic manipulations imply that
y(t) = c e at − b
a .
This establishes the Theorem.
Remarks:
(a) The Proof of Theorem 1.1.2 starts multiplying the differential equation by the exponential
function e−at . People could argue that it is not clear where this idea comes from.
In Lemma 1.1.3 we show that only functions proportional to the exponential e−at have
the property needed to be an integrating factor for the differential equation. The proof
of Lemma 1.1.3 starts multiplying the differential equation by a function µ, and then
one finds that µ(t) = c e−at .
(b) Since the function µ is used to multiply the original differential equation, we can freely
choose the function µ with c = 1. This is the choice made in the proof of Theorem 1.1.2.
(c) It is important we understand the origin of the integrating factor e−at in order to extend
results from constant coefficients equations to variable coefficients equations.
The following Lemma states that those functions proportional to e−at are integrating factors
for the differential equation in (1.1.3).
Lemma 1.1.3 (Integrating Factor). Given any differentiable function y and constant a,
every function µ satisfying
(y ′ − ay)µ = (yµ) ′ ,
must be given by the expression below, for any c ∈ R,
µ(t) = c e −at .
Proof of Lemma 1.1.3: Multiply Eq. (1.1.3) by a non-vanishing otherwise arbitrary
function with values µ(t), and order the terms in the equation as follows
µ(y ′ − a y) = b µ. (1.1.5)
The key idea of the proof is to choose the function µ such that the following equation holds
µ(y ′ − a y) = µy ′ . (1.1.6)
Eq. (1.1.6), this is an equation for µ, since it can be rewritten as follows,
µ y ′ − a µy = µ ′ y + µ y ′ ⇔ −a µy = µ ′ y ⇔ −a µ = µ ′ ,

G. NAGY – ODE July 2, 2013 5
that is, the function y does not appear in Eq. (1.1.6). The equation on the far right above
can be solved for µ as follows:
Integrating in the equation above we obtain
µ ′
µ = −a ⇔ ln(µ) ′ = −a.
ln(µ) = −at + c0 ⇔ e ln(µ) = e −at+c0 = e −at e c0 ,
where c0 is an arbitrary constant. Denoting c = e c0, the integrating factor is the nonvanishing
function
µ(t) = c e −at .
This establishes the Lemma.
We now solve the problem in Example 1.1.3 following the proof of Theorem 1.1.2.
Example 1.1.3: Find all solutions to the constant coefficient equation
Solution: Write down the equation in (1.1.7) as follows,
y ′ = 2y + 3 (1.1.7)
y ′ − 2y = 3.
Multiply this equation by the exponential e −2t , that is,
e −2t y ′ − 2 e −2t y = 3 e −2t ⇔ e −2t y ′ + e −2t ′ y = 3 e −2t .
The equation on the far right above is
(e −2t y) ′ = 3 e −2t .
Integrating on both sides get
(e −2t y) ′
dt = 3 e −2t dt + c,
e −2t y = − 3
2 e−2t + c.
Multiplying the whole equation by e 2t ,
y(t) = c e 2t − 3
, c ∈ R.
2
⊳
y
0
−3/2
c > 0
t
c = 0
c < 0
Figure 1. The graph of a
few solutions to Eq. (1.1.7)
for different constants c.
We now solve same problem above but using the formula in Theorem 1.1.2.
Example 1.1.4: Find all solutions to the constant coefficient equation
y ′ = 2y + 3 (1.1.8)
Solution: The equation above is the particular case of Eq. (1.1.3) given by a = 2 and
b = 3. Therefore, using these values in the expression for the solution given in Eq. (1.1.4)
we obtain
y(t) = ce 2t − 3
2 .
⊳

6 G. NAGY – ODE july 2, 2013
1.1.4. The Initial Value Problem. Sometimes in physics one is not interested in all
solutions to a differential equation, but only in those solutions satisfying an extra condition.
For example, in the case of Newton’s second law of motion for a point particle, one could
be interested only in those solutions satisfying an extra condition: At an initial time the
particle must be at a specified initial position. Such condition is called an initial condition,
and it selects a subset of solutions of the differential equation. An initial value problem
means to find a solution to both a differential equation and an initial condition.
Definition 1.1.4. The initial value problem (IVP) for a constant coefficients first order
linear ODE is the following: Given a, b, t 0, y 0 ∈ R, find a solution y : R → R of the problem
y ′ = a y + b, y(t0) = y0. (1.1.9)
The second equation in (1.1.9) is called the initial condition of the problem. Although
the differential equation in (1.1.9) has infinitely many solutions, the associated initial value
problem has a unique solution.
Theorem 1.1.5 (Constant coefficients IVP). Given constants a, b, t0, y0 ∈ R, with a =
0, the initial value problem in (1.1.9) has the unique solution
y(t) = y0 + b
e
a
a(t−t0) b
− . (1.1.10)
a
To prove this result we use Theorem 1.1.2 to write down the general solution of the
differential equation; the initial condition to fixes the value of the integration constant c.
Proof of Theorem 1.1.5: The general solution of the differential equation in (1.1.9) is
given in Eq. (1.1.4), that is,
y(t) = c e at − b
a .
The initial condition determines the value of the constant c, as follows
y0 = y(t0) = c e at0
b
−
a
⇔
c = y0 + b
e
a
−at0 .
Introduce this expression for the constant c into the differential equation in Eq. (1.1.9),
y(t) = y0 + b
e
a
a(t−t0) b
−
a .
This establishes the Theorem.
Example 1.1.5: Find the unique solution of the initial value problem
y ′ = 2y + 3, y(0) = 1. (1.1.11)
Solution: Every solution of the differential equation is given by y(t) = ce2t − (3/2), where
c is an arbitrary constant. The initial condition in Eq. (1.1.11) determines the value of c,
1 = y(0) = c − 3
2
⇔
5
c =
2 .
Then, the unique solution to the IVP above is
y(t) = 5
2 e2t − 3
2 .
⊳

1.1.5. Exercises.
1.1.1.- Verify that y(t) = (t + 2) e 2t is solution
of the IVP
y ′ = 2y + e 2t , y(0) = 2.
1.1.2.- Find the general solution of
y ′ = −4y + 2
1.1.3.- Find the solution of the IVP
y ′ = −4y + 2, y(0) = 5.
1.1.4.- Find the solution of the IVP
dy
(t) = 3 y(t) − 2, y(1) = 1.
dt
G. NAGY – ODE July 2, 2013 7
1.1.5.- Express the differential equation
y ′ (t) = 6 y(t) + 1 (1.1.12)
as a total derivative of a potential function
ψ(t, y(t)), that is, find ψ satisfying
dψ
= 0
dt
for every y solution of Eq. (1.1.12).
Then, find the general solution y of the
equation above.
1.1.6.- Find the solution of the IVP
y ′ = 6 y + 1, y(0) = 1.

8 G. NAGY – ODE july 2, 2013
1.2. Linear variable coefficients equations
In Section 1.1 we found an expression for the general solution of a first order, linear,
constant coefficient differential equation, Theorem 1.1.2. We also learned that the initial
value problem for such equation has a unique solution, Theorem 1.1.5. In the first part of
this Section we generalize these results to variable coefficients equations, that is, equations
of the form
y ′ (t) = a(t) y(t) + b(t),
where a, b : (t 1, t 2) → R are continuous functions. We will show that the integrating
factor method studied in the last Section can be generalized to variable coefficients linear
equations. In the second part of this Section we introduce the Bernoulli equation, which is a
non-linear differential equation. This non-linear equation has a particular property: It can
be transformed into a linear equation by and appropriate change in the unknown function.
One then solves the linear equation for the changed function using the integrating factor
method. Finally one transforms back the changed function into the original function.
1.2.1. Linear equations with variable coefficients. We start generalizing the result in
Theorem 1.1.2 to variable coefficinents equations.
Theorem 1.2.1 (Variable coefficients). If the functions a, b : (t 1, t 2) → R are continuous,
where t1 < t2, then the differential equation
y ′ = a(t) y + b(t), (1.2.1)
has infinitely many solutions and every solution can be labeled by c ∈ R as follows
y(t) = c e A(t) + e A(t)
e −A(t)
b(t) dt, (1.2.2)
where we introduced the function A(t) = a(t) dt, any primitive of the function a.
Remark: In the particular case of constant coefficients we see that a primitive for the
constant function a ∈ R is A(t) = at, while
e A(t)
e −A(t) b(t) dt = e at
b e −at dt = e at
− b
a e−at = − b
a ,
hence we recover the expression y(t) = c eat − b
given in Eq. (1.1.3).
a
Proof of Theorem 1.2.1: We generalize the integrating factor method to variable coefficients
equations. We first write down the differential equation as
y ′ − a(t) y = b(t).
If we denote any primitive of coefficient a as follows, A(t) = a(t) dt, then multiply the
equation above by the function e −A(t) , called the integrating factor,
y ′ − a(t) y e A(t) = b(t) e −A(t) ⇔ e −A(t) y ′ − a(t) e −A(t) y = b(t) e −A(t) .
This exponential was chosen because of the property
−a(t) e −A(t) = e −A(t) ′ ,
since A ′ (t) = a(t). Introducing this property into the differential equation,
e −A(t) y ′ + e −A(t) ′ y = b(t) e −A(t) .
The product rule for derivatives says that the left-hand side above is a total derivative,
e −A(t) y ′ = b(t) e −A(t) .

G. NAGY – ODE July 2, 2013 9
One way to proceed at this point is to rewrite the right-hand side above in terms of its
primitive function, B(t) = e −A(t) b(t) dt, that is,
e −A(t) y ′ = B ′ (t) ⇔ e −A(t) y − B(t) ′ = 0.
As in the constant coefficient case, the whole differential equation has been rewritten as the
total derivative of a function, in this case,
ψ(t, y(t)) = e −A(t)
y(t) − e −A(t) b(t) dt,
called potential function. The differential equation now has the form
dψ
(t, y(t)) = 0.
dt
It is trivial to integrate the equation in terms of the potential function,
ψ(t, y(t)) = c ⇔ e −A(t)
y(t) − e −A(t) b(t) dt = c.
Therefore, for each value of the constant c ∈ R we have the solution
y(t) = c e A(t) + e A(t)
e −A(t) b(t) dt.
This establishes the Theorem.
Lemma 1.1.3 can be generalized to the variable coefficient case, henceforth stating that
the integrating factor µ(t) = e −A(t) describes all possible integrating factors for Eq. (1.2.1).
Lemma 1.2.2 (Integrating Factor). Given any differentiable function y and integrable
function a, every function µ satisfying
(y ′ − ay)µ = (yµ) ′ ,
must be given in terms of any primitive of function a, A(t) = a(t) dt, as follows,
µ(t) = e −A(t) .
Proof of Lemma 1.2.2: Multiply Eq. (1.2.1) by a non-vanishing otherwise arbitrary
function with values µ(t), and order the terms in the equation as follows
µ(y ′ − a(t) y) = b(t) µ. (1.2.3)
The key idea of the proof is to choose the function µ such that the following equation holds
µ(y ′ − a(t) y) = µy ′ . (1.2.4)
Eq. (1.2.4), this is an equation for µ, since it can be rewritten as follows,
µ y ′ − a(t) µy = µ ′ y + µ y ′ ⇔ −a(t) µy = µ ′ y ⇔ −a(t) µ = µ ′ ,
that is, the function y does not appear in Eq. (1.2.4). The equation on the far right above
can be solved for µ as follows:
µ ′
µ = −a(t) ⇔ ln(µ) ′ = −a(t).
Integrating in the equation above, and introducing a primitive A(t) = a(t) dt, we obtain
ln(µ) = −A(t) ⇔ e ln(µ) = e −A(t) ⇔ µ(t) = e −A(t) .
This establishes the Lemma.

10 G. NAGY – ODE july 2, 2013
Example 1.2.1: Find all solutions y to the differential equation
y ′ = 3
t y + t5 .
Solution: Rewrite the equation as
y ′ − 3
t y = t5 .
Introduce a primitive of the coefficient function a(t) = 3/t,
3
A(t) =
t dt = 3 ln(t) = ln(t3 ),
so we have A(t) = ln(t 3 ). The integrating factor µ is then
µ(t) = e −A(t) = e − ln(t3 ) = e ln(t −3 ) = t −3 ,
hence µ(t) = t −3 . We multiply the differential equation by the integrating factor
t −3 y ′ − 3
t y − t5 = 0 ⇔ t −3 y ′ − 3 t −4 y − t 2 Since −3 t
= 0.
−4 = (t−3 ) ′ , we get
t −3 y ′ + (t −3 ) ′ y − t 2 = 0 ⇔
t −3 y − t3
3
′
= 0.
The potential function in this case is ψ(t, y) = t −3 y − t3
. Integrating the total derivative
3
we obtain
t −3 y − t3
3 = c ⇒ t−3 y = c + t3
3 ,
so all solutions to the differential equation are y(t) = c t 3 + t6
, with c ∈ R. ⊳
3
1.2.2. The Initial Value Problem. We now generalize to variable coefficients equations
the existence and uniqueness of solutions to initial value problems for constant coefficients
linear differential equations presented in Theorems 1.1.5. We first introduce the initial value
problem for a variable coefficients equations, a simple generalization of Def. 1.1.4.
Definition 1.2.3. The initial value problem (IVP) for a first order linear ODE is the
following: Given functions a, b : R → R and constants t0, y0 ∈ R, find y : R → R solution of
y ′ = a(t) y + b(t), y(t 0) = y 0. (1.2.5)
As we did with the constant coefficients IVP, the second equation in (1.2.5) is called the
initial condition of the problem. We saw in Theorem 1.2.1 that the differential equation
in (1.2.5) has infinitely many solutions, parametrized by a real constant c. The associated
initial value problem has a unique solution though, because the initial condition fixes the
arbitrary ccnstant c.
Theorem 1.2.4 (Variable coefficients IVP). Given continuous functions a, b : (t 1, t 2) →
R and constants t0 ∈ (t1, t2) and y0 ∈ R, the initial value problem
has the unique solution y : (t 1, t 2) → R given by
y ′ = a(t) y + b(t), y(t 0) = y 0, (1.2.6)
y(t) = e A(t)
y0 +
t
t0
e −A(s)
b(s) ds , (1.2.7)

where we introduced the function A(t) =
G. NAGY – ODE July 2, 2013 11
t
t0
a(s) ds, a particular primitive of function a.
Remark: In the particular case of a constant coefficients equation, that is, a, b ∈ R, the
solution given in Eq. (1.2.7) reduces to the one given in Eq. (1.1.10). Indeed,
A(t) = −
t
t0
a ds = −a(t − t 0),
t
t0
e −a(s−t0) b ds = − b
a e−a(t−t0) + b
a .
Therefore, the solution y can be written as
y(t) = y0 e a(t−t0)
+ − b
a e−a(t−t0) b
+ e
a
a(t−t0)
= y0 + b
e
a
a(t−t0) b
−
a .
Proof Theorem 1.2.4: We follow closely the proof of Theorem 1.1.2. From Theorem 1.2.1
we know that all solutions to the differential equation in (1.2.6) are given by
y(t) = c e A(t) + e A(t)
e −A(t) b(t) dt,
for every c ∈ R. Let us use again the notation B(t) = e −A(t) b(t) dt, and then introduce
the initial condition in (1.2.6), which fixes the constant c,
so we get the constant c,
y0 = y(t0) = c e A(t0) + e A(t0) B(t0),
c = y0 e −A(t0) − B(t0).
Using this expression in the general solution above,
y(t) = y0 e −A(t0)
− B(t0) e A(t) + e A(t) B(t) = y0 e A(t)−A(t0) A(t)
+ e B(t) − B(t0) .
Let us introduce the particular primitives Â(t) = A(t) − A(t0) and ˆ B(t) = B(t) − B(t0),
which vanish at t0, that is,
Â(t) =
t
t0
a(s) ds,
Then the solution y of the IVP has the form
which is equivalent to
so we conclude that
ˆ B(t) =
y(t) = y0 e Â(t) + e A(t)
t
y(t) = y0 e Â(t) + e A(t)−A(t0)
t
y(t) = e Â(t)
y 0 +
t
t0
t0
t0
t
t0
e −A(s) b(s) ds.
e −A(s) b(s) ds
e −[A(s)−A(t0)] b(s) ds,
e −Â(s)
b(s) ds .
Renaming the particular primitive  simply by A, we then establish the Theorem.
We solve the next Example following the main steps in the proof of Theorem 1.2.4 above.
Example 1.2.2: Find the function y solution of the initial value problem
ty ′ + 2y = 4t 2 , y(1) = 2.

12 G. NAGY – ODE july 2, 2013
Solution: We first write the equation above in a way it is simple to see the functions a
and b in Theorem 1.2.4. In this case we obtain
y ′ = − 2
t
Now rewrite the equation as
and multiply it by the function µ = e A(t) ,
y + 4t ⇔ a(t) = −2 , b(t) = 4t. (1.2.8)
t
y ′ = 2
y = 4t,
t
e A(t) y ′ + 2
t eA(t) y = 4t e A(t) .
The function A in the integrating factor e A(t) must be the one satisfying
A ′ (t) = 2
t
⇔ A(t) =
In this case the differential equation can be written as
2
t dt.
e A(t) y ′ + A ′ (t) e A(t) y = 4t e A(t) ⇔ e A(t) y ′ = 4t e A(t) .
We now compute the function A,
dt
A(t) = 2
t = 2 ln(|t|) ⇒ A(t) = ln(t2 ).
This implies that
e A(t) = t 2 .
Therefore, the differential equation can be written as
t 2 y ′ = 4t 3 .
Integrating on both sides we obtain that
The initial condition implies that
t 2 y = t 4 + c ⇒ y(t) = t 2 + c
.
t2 2 = y(1) = c + 1 ⇒ c = 1 ⇒ y(t) = 1
t 2 + t2 .
Remark: It is not needed to compute the potential function to find the solution in the
Example above. However, it could be useful to see this function for the differential equation
in the Example. When the equation is written as
t 2 y ′ = 4t 3 ⇔ (t 2 y) ′ = (t 4 ) ′ ⇔ (t 2 y − t 4 ) ′ = 0,
it is simple to see that the potential function is
ψ(t, y(t)) = t 2 y(t) − t 4 .
The differential equation is then trivial to integrate,
t 2 y − t 4 = c ⇔ t 2 y = c + t 4 ⇔ y(t) = c
t 2 + t2 .
⊳

G. NAGY – ODE July 2, 2013 13
Example 1.2.3: Find the solution of the problem given in Example 1.2.2 using the results
of Theorem 1.2.4.
Solution: We find the solution simply by using Eq. (1.2.7). First, find the integrating
factor function µ as follows:
A(t) = −
t
1
2
s ds = −2 ln(t) − ln(1) = −2 ln(t) ⇒ A(t) = ln(t −2 ).
The integrating factor is µ(t) = e −A(t) , that is,
Then, we compute the solution as follows:
µ(t) = e − ln(t−2 ) = e ln(t 2 ) ⇒ µ(t) = t 2 .
y(t) = 1
t2
2 +
= 2 1
+
t2 t2 2
1
t
1
s 2
4s ds
4s 3 ds
= 2 1
+
t2 t2 (t4 − 1)
= 2
t2 + t2 − 1
t2 ⇒
1
y(t) =
t2 + t2 .
1.2.3. The Bernoulli equation. In 1696 Jacob Bernoulli struggled for months trying to
solve a particular differential equation, now known as Bernoulli’s differential equation, see
Def. 1.2.5. He could not solve it, so he organized a contest among his peers to solve the
equation. In short time his brother Johann Bernoulli solved it. Later on the equation was
solved by Leibniz, transforming the original non-linear equation into a linear equation. In
what follows we explain Leibniz’s idea in more detail.
Definition 1.2.5. A Bernoulli equation in the unknown function y, determined by the
functions p, q : (t1, t2) → R and a number n ∈ R, is the differential equation
y ′ = p(t) y + q(t) y n . (1.2.9)
In the case that n = 0 or n = 1 the Bernoulli equation reduces to a linear equation. The
interesting cases are when the Bernoulli equation is non-linear. We now show in an Example
the main idea to solve a Bernoulli equation: To transform the original non-linear equation
into a linear equation.
Example 1.2.4: Find every solution of the differential equation
y ′ = y + 2y 5 .
Solution: This is a Bernoulli equation for n = 5. Divide the equation by y 5 ,
y ′ 1
= + 2.
y5 y4 Introduce the function v = 1/y 4 and its derivative v ′ = −4(y ′ /y 5 ), into the differential
equation above,
− v′
4 = v + 2 ⇒ v′ = −4 v − 8 ⇒ v ′ + 4 v = −8.
⊳

14 G. NAGY – ODE july 2, 2013
The last equation is a linear differential equation for v. This equation can be solved using
the integrating factor method. Multiply the equation by µ(t) = e 4t , then
e 4t v ′ = −8 e 4t ⇒ e 4t v = − 8
4 e4t + c
We obtain that v = c e −4t − 2. Since v = 1/y 4 ,
1
y4 = c e−4t − 2 ⇒ y(t) = ±
1
c e −4t − 2 1/4 .
The following result summarizes the main ideas used in the previous example to find
solutions to the non-linear Bernoulli equation.
Theorem 1.2.6 (Bernoulli). The function y is a solution of the Bernoulli equation
y ′ = p(t) y + q(t) y n , n = 1,
iff the function v = 1/y (n−1) is solution of the linear differential equation
v ′ = −(n − 1)p(t) v − (n − 1)q(t).
This result says, solve the linear equation for function v, a the compute the solution y of
the Bernoulli equation as y = (1/v) 1/(n−1) .
Proof of Theorem 1.2.6: Divide the Bernoulli equation by y n ,
y ′ p(t)
= + q(t).
yn yn−1 Introduce the new unknown v = y −(n−1) and compute its derivative,
v ′ = y −(n−1) ′ = −(n − 1)y −n y ′ ⇒ − v ′ (t)
(n − 1) = y′ (t)
y n (t) .
If we substitute v and this last equation into the Bernoulli equation we get
− v′
(n − 1) = p(t) v + q(t) ⇒ v′ = −(n − 1)p(t) v − (n − 1)q(t).
This establishes the Theorem.
Example 1.2.5: Given any constants a0, b0, find every solution of the differential equation
y ′ = a0y + b0y 3 .
Solution: This is a Bernoulli equation. Divide the equation by y 3 ,
y ′
y
3 = a0
+ b0.
y2 Introduce the function v = 1/y 2 and its derivative v ′ = −2(y ′ /y 3 ), into the differential
equation above,
− v′
2 = a0v + b0 ⇒ v ′ = −2a0v − 2b0 ⇒ v ′ + 2a0v = −2b0.
The last equation is a linear differential equation for v. This equation can be solved using
the integrating factor method. Multiply the equation by µ(t) = e2a0t ,
2a0t ′
e v = −2b0 e 2a0t ⇒ e 2a0t v = − b0
a0
e 2a0t + c
⊳

We obtain that v = c e −2a0t − b0
. Since v = 1/y2 ,
a0
1
y 2 = c e−2a0t − b0
a0
G. NAGY – ODE July 2, 2013 15
⇒ y(t) = ±
1
.
c e−2a0t b0
1/2
− a0
Example 1.2.6: Find every solution of the equation t y ′ = 3y + t 5 y 1/3 .
Solution: Rewrite the differential equation as
y ′ = 3
t y + t4 y 1/3 .
This is a Bernoulli equation for n = 1/3. Divide the equation by y 1/3 ,
y ′ 3
=
y1/3 t y2/3 + t 4 .
Define the new unknown function v = 1/y (n−1) , that is, v = y 2/3 , compute is derivative,
v ′ = 2
3
y ′
, and introduce them in the differential equation,
y1/3 3
2 v′ = 3
t v + t4 ⇒ v ′ − 2 2
v =
t 3 t4 .
Integrate the linear equation of v using the integrating factor method.
factor is computed as follows:
2
A(t) =
t
The integrating
dt = 2 ln(t) = ln(t2 ),
and recalling that µ(t) = e −A(t) , we obtain
µ(t) = e − ln(t2 ) ln(t
= e −2 ) 1
⇒ µ(t) = . 2
t
Therefore, the equation for v is
1
t2 ′ 2
v −
t v = 2
3 t2
v 2
⇒ −
t2 9 t3 ′
= 0.
The potential function is ψ(t, v) = v/t2−(2/9)t3 and the solution of the differential equation
is ψ(t, v(t)) = c, that is,
v 2
−
t2 9 t3
2
= c ⇒ v(t) = t c + 2
9 t3 ⇒ v(t) = c t 2 + 2
9 t5 .
Once v is known we compute the original unknown y = ±v3/2 , where the double sign is
related to taking the square root. So we conclude,
y(t) = ± c t 2 + 2
9 t5 3/2
⊳
⊳

16 G. NAGY – ODE july 2, 2013
1.2.4. Exercises.
1.2.1.- Find the solution y to the IVP
y ′ = −y + e −2t , y(0) = 3.
1.2.2.- Find the solution y to the IVP
y ′ = y + 2te 2t , y(0) = 0.
1.2.3.- Find the solution y to the IVP
t y ′ + 2 y = sin(t)
π
, y =
t 2
2
π ,
for t > 0.
1.2.4.- Find all solutions y to the ODE
y ′
(t2 = 4t.
+ 1)y
1.2.5.- Find all solutions y to the ODE
ty ′ + n y = t 2 ,
with n a positive integer.
1.2.6.- Find all solutions to the ODE
2ty − y ′ = 0.
Show that given two solutions y1 and
y2 of the equation above, the addition
y1 + y2 is also a solution.
1.2.7.- Find every solution of the equation
y ′ + t y = t y 2 .
1.2.8.- Find every solution of the equation
y ′ = −x y = 6x √ y.

G. NAGY – ODE July 2, 2013 17
1.3. Separable equations
1.3.1. Separable equations. Non-linear differential equations in general are more complicated
to solve than the linear ones. However, one particular type of non-linear differential
equations are among the simplest equations to solve, even simpler than linear differential
equations. This type of non-linear equations is called separable equations, and they are
solved simply by integrating on both sides of the differential equation.
Definition 1.3.1. Given functions h, g : R → R, a first order ODE on the unknown function
y : R → R is called separable iff the ODE has the form
h(y) y ′ (t) = g(t).
It is not difficult to see that a differential equation y ′ (t) = f(t, y(t)) is separable iff
Example 1.3.1:
(a) The differential equation
y ′ = g(t)
h(y)
is separable, since it is equivalent to
(b) The differential equation
y ′ (t) =
1 − y 2 y ′ (t) = t 2 ⇒
is separable, since it is equivalent to
⇔ f(t, y) = g(t)
h(y) .
t 2
1 − y 2 (t)
g(t) = t 2 ,
y ′ (t) + y 2 (t) cos(2t) = 0
1
y 2 y′ (t) = − cos(2t) ⇒
⎧
⎨
⎩
h(y) = 1 − y 2 .
g(t) = − cos(2t),
h(y) = 1
.
y2 The functions g and h are not uniquely defined; another choice in this example is:
g(t) = cos(2t), h(y) = − 1
.
y2 (c) The linear differential equation y ′ (t) = −a(t) y(t) is separable, since it is equivalent to
1
y y′ (t) = −a(t) ⇒
⎧
⎨
⎩
g(t) = −a(t),
h(y) = 1
y .
(d) The constant coefficients linear differential equation y ′ (t) = −a0 y(t) + b0 is separable,
since it is equivalent to
1
(−a0 y + b0) y′ (t) = 1 ⇒
⎧
⎨ g(t) = 1,
1
⎩ h(y) =
(−a0 y + b0) .
(e) The differential equation y ′ (t) = e y(t) + cos(t) is not separable.
(f) The linear differential equation y ′ (t) = −a(t) y(t) + b(t), with b(t) non-constant, is not
separable.

18 G. NAGY – ODE july 2, 2013
We see in the examples above that a linear ODE with function b = 0 is separable, so the
solutions of such equation can be computed using both the result we show below and the
integrating factor method studied in Sect. 1.2. We now describe a way to find solutions to
every first order separable differential equation.
Theorem 1.3.2 (Separable equations). If the functions h, g : R → R are continuous,
with h = 0 and with primitives H and G, respectively; that is,
then, the separable ODE
H ′ (u) = h(u), G ′ (t) = g(t),
h(y) y ′ = g(t) (1.3.1)
has infinitely many solutions y : R → R satisfying the algebraic equation
where c ∈ R is arbitrary.
H(y(t)) = G(t) + c, (1.3.2)
Before presenting the proof of the Theorem above we use it in the example below and we
explicitly show how to find the primitives G and H of the given functions g and h.
Example 1.3.2: Find all solutions y : R → R to the ODE
y ′ (t) =
t2 1 − y2 . (1.3.3)
(t)
Solution: Theorem 1.3.2 tell us how to obtain the solution y. Writing Eq. (1.3.3) as
1 − y 2 y ′ (t) = t 2 ,
we see that the functions h, g are given by
h(u) = 1 − u 2 , g(t) = t 2 .
Their primitive functions, H and G, respectively, are simple to compute,
h(u) = 1 − u 2 ⇒ H(u) = u − u3
3 ,
g(t) = t 2 ⇒ G(t) = t3
3 .
Then, Theorem (1.3.2) implies that the solution y satisfies the algebraic equation
y(t) − y3 (t) t3
= + c,
3 3
(1.3.4)
where c ∈ R is arbitrary. ⊳
Notice that we have not obtained a definition of a solution function y in Example 1.3.2
above. We have obtained an algebraic equation satisfied by the solution, Eq. (1.3.4). The
equation is algebraic, since it does not contain any derivatives of function y. This is precisely
the result of Theorem 1.3.2 given in Eq. (1.3.2), which is an algebraic equation satisfied by
the solution y. The following notions are useful in this Section.
Definition 1.3.3. Assume the notation in Theorem 1.3.2. The solution y of a separable
ODE is given in implicit form iff function y is specified by
H y(t) = G(t) + c,
⊳

G. NAGY – ODE July 2, 2013 19
The solution y of a separable ODE is given in explicit form iff function H is invertible
and y is specified by
y(t) = H −1 G(t) + c .
Sometimes is difficult to find the inverse of function H. This is the case in Example 1.3.2.
In such cases we leave the solution y written in implicit form. If H−1 is simple to compute,
we express the solution y in explicit form. We now show the proof of Theorem 1.3.2, which
is based in an integration by substitution.
Proof of Theorem 1.3.2: Integrate on both sides in (1.3.1),
h(y(t)) y ′
(t) = g(t) ⇒ h(y(t)) y ′
(t) dt = g(t) dt + c,
where c is an arbitrary constant. Introduce on the left-hand side of the second equation
above the substitution
u = y(t), du = y ′ (t) dt,
which implies,
h(y(t)) y ′
(t) dt = h(u)du ⇒ h(u) du = g(t) dt + c.
Recalling that H and G are the primitives of h and g, respectively,
H(u) = h(u) du, G(t) = g(t) dt,
then we obtain that
H(u) = G(t) + c.
Substitute y(t) back in the place of u. We conclude that the solution function y satisfies
the algebraic equation
H y(t) = G(t) + c.
This establishes the Theorem.
In the following Example we show that an initial value problem can be solved even in the
case where the solutions of the ODE are given only in implicit form.
Example 1.3.3: Find the solution of the initial value problem
y ′ (t) =
t2 1 − y2 , y(0) = 1. (1.3.5)
(t)
Solution: From Example 1.3.2 we know that all solutions to the differential equation
in (1.3.5) are given by
y(t) − y3 (t) t3
= + c,
3 3
where c ∈ R is arbitrary. This constant c is now fixed with the initial condition in Eq. (1.3.5)
y(0) − y3 (0)
1
2
= c ⇒ 1 − = c ⇔ c =
3 3 3 ⇒ y(t) − y3 (t) t3 2
= +
3 3 3 .
So we can rewrite the algebraic equation defining the solution functions y as the roots of a
cubic polynomial,
y 3 (t) − 3y(t) + t 3 + 2 = 0.
⊳
In Examples 1.3.2 and 1.3.3 above we have used Theorem 1.3.2 to find the implicit form
of the solution y. In the Example below we solve the same problem than in Example 1.3.2
but now we follow step by step the proof of Theorem 1.3.2.

20 G. NAGY – ODE july 2, 2013
Example 1.3.4: Follow the proof of Theorem 1.3.2 to find all solutions y to the equation
y ′ (t) =
Solution: We write the differential equation in (1.3.6) as follows
1 − y 2 (t) y ′ (t) = t 2 ;
t2 1 − y2 . (1.3.6)
(t)
We now integrate in the variable t on both sides of the equation above,
1 2 ′
− y (t) y (t) dt = t 2 dt + c;
where c is any constant, since the integrals are indefinite. Introduce the substitution
then we obtain
(1 − u 2
) du =
u = y(t), du = y ′ (t) dt,
t 2 dt + c ⇔ u − u3
3
= t3
3
Substitute back the original unknown y into the last expression above and we obtain
y(t) − y3 (t) t3
= + c,
3 3
which is the implicit form of the solution y. ⊳
We present now few more Examples.
Example 1.3.5: Find the solution of the initial value problem
+ c.
y ′ (t) + y 2 (t) cos(2t) = 0, y(0) = 1. (1.3.7)
Solution: The differential equation above is separable, with
g(t) = − cos(2t), h(y) = 1
,
y2 therefore, it can be integrated as follows:
Again the substitution
y ′ (t)
y2 = − cos(2t) ⇔
(t)
′ y (t)
y2
dt = −
(t)
u = y(t), du = y ′ (t) dt
cos(2t) dt + c.
implies that
du
= − cos(2t) dt + c ⇔ −
u2 1
= −1 sin(2t) + c.
u 2
Substitute the unknown function y back in the equation above,
− 1
= −1 sin(2t) + c.
y(t) 2
The solution is given in implicit form. However, in this case is simple to solve this algebraic
equation for y and we obtain the following explicit form for the solutions,
2
y(t) =
sin(2t) − 2c .
The initial condition implies that
1 = y(0) = 2
0 − 2c
⇔ c = −1.

So, the solution to the IVP is given in explicit form by
G. NAGY – ODE July 2, 2013 21
y(t) =
2
sin(2t) + 2 .
Example 1.3.6: Follow the proof in Theorem 1.3.2 to find all solutions y of the ODE
y ′ (t) =
4t − t3
4 + y 3 (t) .
Solution: The differential equation above is separable, with
therefore, it can be integrated as follows:
4 + y 3 (t) y ′ (t) = 4t − t 3 ⇔
Again the substitution
g(t) = 4t − t 3 , h(y) = 4 + y 3 ,
4 + y 3 (t) y ′ (t) dt =
u = y(t), du = y ′ (t) dt
(4t − t 3 ) dt + c.
implies that
(4 + u 3
) du = (4t − t 3 ) dt + c0. ⇔ 4u + u4
4 = 2t2 − t4
4 + c0. Substitute the unknown function y back in the equation above and calling c1 = 4c0 we
obtain the following implicit form for the solution,
y 4 (t) + 16y(t) − 8t 2 + t 4 = c 1.
Example 1.3.7: Find the explicit form of the solution to the initial value problem
y ′ (t) =
2 − t
1 + y(t)
Solution: The differential equation above is separable with
Their primitives are respectively given by,
g(t) = 2 − t, h(u) = 1 + u.
y(0) = 1. (1.3.8)
g(t) = 2 − t ⇒ G(t) = 2t − t2
, h(u) = 1 + u
2
⇒
u2
H(u) = u +
2 .
Therefore, the implicit form of all solutions y to the ODE above are given by
y(t) + y2 (t)
2
= 2t − t2
2
with c ∈ R. The initial condition in Eq. (1.3.8) fixes the value of constant c, as follows,
+ c,
y(o) + y2 (0)
= 0 + c
2
⇒
1
1 + = c
2
⇒
3
c =
2 .
We conclude that the implicit form of the solution y is given by
y(t) + y2 (t)
2
= 2t − t2
2
+ 3
2 , ⇔ y2 (t) + 2y(t) + (t 2 − 4t − 3) = 0.
⊳
⊳

22 G. NAGY – ODE july 2, 2013
The explicit form of the solution can be obtained realizing that y(t) is a root in the quadratic
polynomial above. The two roots of that polynomial are given by
y±(t) = 1
−2 ± 4 − 4(t2 − 4t − 3) ⇔ y±(t) = −1 ±
2
−t2 + 4t + 4.
We have obtained two functions y+ and Y−. However, we know that there is only one
solution to the IVP. We can decide which one is the solution by evaluating them at the
value t = 0 given in the initial condition. We obtain
y+(0) = −1 + √ 4 = 1,
y−(0) = −1 − √ 4 = −3.
Therefore, the solution is y+, that is, the explicit form of the solution is
y(t) = −1 + −t 2 + 4t + 4.
1.3.2. Euler Homogeneous equations. Sometimes a differential equation is not separable
but a change of unknown in the equation can transform a non-separable equation into
a separable equation. This is the case for a type of differential equations called Euler
homogeneous equations.
Definition 1.3.4. The first order differential equation of the form y ′ (t) = f t, y(t) is called
Euler homogeneous iff for every real numbers t, u and every c = 0 the function f satisfies
f(ct, cu) = f(t, u).
A differential equation is Euler homogeneous when the function f is invariant under the
change of scale of its arguments. It is not difficult to prove that a function with values
f(t, u) having the property above must depend on (t, u) only through their quotient. That
is, there exists a function F : R → R such that
u
f(t, u) = F .
t
Therefore, a first order differential equation is Euler homogeneous iff it has the form
y ′ (t) = F
y(t)
t
Example 1.3.8: Show that the equation below is Euler homogeneous,
.
(t − y) y ′ − 2y + 3t + y2
t
Solution: Rewrite the equation in the standard form
(t − y) y ′ = 2y − 3t − y2
t ⇒ y′ =
= 0.
2y − 3t − y2
t .
(t − y)
Divide numerator and denominator by t. We get,
y ′
2y − 3t −
=
y2
1
t
t
(t − y) 1
t
⇒ y ′
y
y
2 2 − 3 −
= t
t
y
.
1 −
t
⊳

G. NAGY – ODE July 2, 2013 23
We conclude that the differential equation is Euler homogeneous, because the right-hand side
of the equation above depends only on y/t. Indeed, we have shown that f(t, y) = F (y/t),
where
f(t, y) = 2y − 3t − (y2 /t)
, F (x) =
t − y
2x − 3 − x2
,
1 − x
Example 1.3.9: Determine whether the equation below is Euler homogeneous,
y ′ = t2
.
1 − y3 Solution: Dividing numerator and denominator by t3 we obtain
y ′ t
=
2
(1 − y3
1
t
)
3
1
t3 ⇒ y ′
1
= t
1
t3
y
3 .
−
t
We conclude that the differential equation is not Euler homogeneous. ⊳
Theorem 1.3.5 (Euler Homogeneous). If the differential equation y ′ (t) = f t, y(t) is
Euler homogeneous, then the differential equation for the unknown v(t) = y(t)
is separable.
t
The Theorem says that Euler homogeneous equations can be transformed into separable
equations. This is analogous to the idea used to solve a Bernoulli equation, where we
transformed the non-linear equation into a linear one. In the case of an Euler homogeneous
equation for the unknown function y, we transform it into a separable equation for the
unknown function v = y/t. We then solve for v in implicit or explicit form. Finally we
replace back y = t v.
Proof of Theorem 1.3.5: If y ′ = f(t, y) is homogeneous, then it can be written as
y ′ = F (y/t) for some function F . Introduce v = y/t. This means,
y(t) = t v(t) ⇒ y ′ (t) = v(t) + t v ′ (t).
Introducing these expressions into the differential equation for y we get
F (v) − v
v + t v ′ = F (v) ⇒ v ′ =
This last equation is separable. This establishes the Theorem.
Example 1.3.10: Find all solutions y of the ODE y ′ = t2 + 3y2 .
2ty
Solution: The equation is homogeneous, since
y ′ = (t2 + 3y2
1
) t
2ty
2
1
t2 ⇒ y ′
y
2 1 + 3
= t
y
.
2
t
Therefore, we introduce the change of unknown v = y/t, so y = t v and y ′ = v + t v ′ . Hence
v + t v ′ =
1 + 3v2
2v
⇒ t v ′ =
t
1 + 3v2
2v − v = 1 + 3v2 − 2v2 2v
.
⊳

24 G. NAGY – ODE july 2, 2013
We obtain the separable equation v ′ = 1
2 1 + v
. We rewrite and integrate it,
t 2v
2v
1 + v2 v′ = 1
t ⇒
2v
1 + v2 v′
1
dt = dt + c0.
t
The substitution u = 1 + v 2 (t) implies du = 2v(t) v ′ (t) dt, so
du
u =
dt
t + c0 ⇒ ln(u) = ln(t) + c0 ⇒ u = e ln(t)+c0 .
But u = e ln(t) e c0 , so denoting c1 = e c0 , then u = c1t. Hence, the explicit form of the solution
can be computed as follows,
1 + v 2 = c1t ⇒ 1 +
y
2 = c1t ⇒ y(t) = ±t
t
√ c1t − 1.
Example 1.3.11: Find all solutions y of the ODE y ′ t(y + 1) + (y + 1)2
= .
Solution: This equation is not homogeneous in the unknown y and variable t, however,
it becomes homogeneous in the unknown u(t) = y(t) + 1 and the same variable t. Indeed,
u ′ = y ′ , thus we obtain
y ′ t(y + 1) + (y + 1)2
=
t2 ⇔ u ′ tu + u2
=
t2 ⇔ u ′ = u
t +
t
Therefore, we introduce the new variable v = u/t, which satisfies u = t v and u ′ = v + t v ′ .
The differential equation for v is
v + t v ′ = v + v 2 ⇔ t v ′ = v 2 ⇔
v ′
t 2
dt =
v2 u
1
dt + c,
t
with c ∈ R. The substitution w = v(t) implies dw = v ′ dt, so
w −2
1
dw =
t dt + c ⇔ −w−1 1
= ln(|t|) + c ⇔ w = −
ln(|t|) + c .
Substituting back v, u and y, we obtain w = v(t) = u(t)/t = [y(t) + 1]/t, so
y + 1 1
= −
t ln(|t|) + c
⇔
t
y(t) = − − 1.
ln(|t|) + c
2
.
⊳
⊳

1.3.3. Exercises.
1.3.1.- Find all solutions y to the ODE
y ′ = t2
y .
Express the solutions in explicit form.
1.3.2.- Find every solution y of the ODE
3t 2 + 4y 3 y ′ − 1 + y ′ = 0.
Leave the solution in implicit form.
1.3.3.- Find the solution y to the IVP
y ′ = t 2 y 2 , y(0) = 1.
1.3.4.- Find every solution y of the ODE
ty + 1 + t 2 y ′ = 0.
G. NAGY – ODE July 2, 2013 25
1.3.5.- Find every solution y of the Euler
homogeneous equation
y ′ y + t
= .
t
1.3.6.- Find all solutions y to the ODE
y ′ = t2 + y 2
.
ty
1.3.7.- Find the explicit solution to the IVP
(t 2 + 2ty) y ′ = y 2 , y(1) = 1.
1.3.8.- Prove that if y ′ = f(t, y) is an Euler
homogeneous equation and y1(t) is a solution,
then y(t) = (1/k) y1(kt) is also a
solution for every non-zero k ∈ R.

26 G. NAGY – ODE july 2, 2013
1.4. Exact equations
We introduce a particular type of differential equations called exact. They are called exact
differential equations because they can be rewritten as the vanishing of a total derivative
of an appropriate function, called potential function. This property makes it simple to find
implicit solutions to an exact equation. We just need to find this potential function. In
the second part of this Section we study a type of differential equations that are not exact
but they can be converted into exact equations when they are multiplied by an appropriate
function. This function is called an integrating factor. The integrating factor converts a
non-exact equation into an exact equation. Linear differential equations are a particular
case of this type of equations, and we have studied them in Sections 1.1 and 1.2. For linear
equations we computed integrating factors that transformed the equation into a derivative
of a potential function. We now generalize this idea to a class of non-linear equations.
1.4.1. Exact differential equations. We start with a definition of an exact equation.
Definition 1.4.1. The differential equation in the unknown function y : (t1, t2) → R
N(t, y(t)) y ′ (t) + M(t, y(t)) = 0
is called exact in an open rectangle R = (t1, t2) × (u1, u2) ⊂ R 2 iff for every point (t, u) ∈ R
the functions M, N : R → R are continuously differentiable and satisfy the equation
∂tN(t, u) = ∂uM(t, u)
We use the notation for partial derivatives ∂tN = ∂N
∂t and ∂uM = ∂M
. Let us see
∂u
whether the following equations are exact or not.
Example 1.4.1: Show whether the differential equation below is exact or not,
2ty(t) y ′ (t) + 2t + y 2 (t) = 0.
Solution: We first identify the functions N and M. This is simple in this case, since
2ty(t) y ′ (t) + 2t + y 2 (t) = 0 ⇒ N(t, u) = 2tu, M(t, u) = 2t + u 2 .
The equation is indeed exact, since
N(t, u) = 2tu ⇒ ∂tN(t, u) = 2u,
M(t, u) = 2t + u 2 ⇒ ∂uM(t, u) = 2u,
⇒ ∂tN(t, u) = ∂uM(t, u).
Example 1.4.2: Show whether the differential equation below is exact or not,
sin(t)y ′ (t) + t 2 e y(t) y ′ (t) − y ′ (t) = −y(t) cos(t) − 2te y(t) .
Solution: We first identify the functions N and M by rewriting the equation as follows,
sin(t) + t 2 e y(t) − 1 y ′ (t) + y(t) cos(t) + 2te y(t) = 0
we can see that
N(t, u) = sin(t) + t 2 e u − 1 ⇒ ∂tN(t, u) = cos(t) + 2te u ,
M(t, u) = u cos(t) + 2te u
Therefore, the equation is exact, since
⇒ ∂uM(t, u) = cos(t) + 2te u .
∂tN(t, u) = ∂uM(t, u).
⊳
⊳

G. NAGY – ODE July 2, 2013 27
The last example shows whether the linear ODE we studied in Section 1.2 is exact.
Example 1.4.3: Show whether the linear differential equation below is exact or not,
y ′ (t) = −a(t) y(t) + b(t), a(t) = 0.
Solution: We first find the functions N and M rewriting the equation as follows,
y ′ + a(t)y − b(t) = 0 ⇒ N(t, u) = 1, M(t, u) = a(t) u − b(t).
We can see that the differential equation is not exact, since
N(t, u) = 1 ⇒ ∂tN(t, u) = 0,
⇒ ∂tN(t, u) = ∂uM(t, u).
M(t, u) = a(t)u − b(t) ⇒ ∂uM(t, u) = a(t),
1.4.2. The Poincaré Lemma. It is well-known how to find implicit solutions to exact
differential equations. The main idea is to find the potential function mentioned at the
begining of this Section. We first show that this potential function always exists for an
exact equation. This result is due to Henri Poincaré around 1880 and we show it without
proof. Later on we show a method to find the potential function whose existence is asserted
by Poincaré’s result.
Lemma 1.4.2 (Poincaré). Continuously differentiable functions M, N : R → R, on an
open rectangle R = (t1, t2) × (u1, u2), satisfy the equation
∂tN(t, u) = ∂uM(t, u) (1.4.1)
iff there exists a twice continuously differentiable function ψ : R → R, called potential
function, such that for all (t, u) ∈ R holds
∂uψ(t, u) = N(t, u), ∂tψ(t, u) = M(t, u). (1.4.2)
Proof of Lemma 1.4.2:
(⇐) We assume that the potential function ψ is given ad satisfies Eq. (1.4.2). Since ψ is
twice continuously differentiable, its cross derivatives are the same, that is, ∂t∂uψ = ∂u∂tψ.
We then conclude that
∂tN = ∂t∂uψ = ∂u∂tψ = ∂uM.
(⇒) It is not given. See [2].
We now show the potential function for a particular exact differential equation. Later on
we show how to actually find such potential function.
Example 1.4.4: Show that the function ψ(t, u) = t 2 + tu 2 is the potential function for the
exact differential equation
2ty(t) y ′ (t) + 2t + y 2 (t) = 0.
Solution: We saw in Example 1.4.1 that the differential equation above is exact, since the
functions M and N,
N(t, u) = 2tu, M(t, u) = 2t + u 2 ,
satisfy that ∂tN = 2u = ∂uM. It is simple to see that ψ(t, u) = t 2 + tu 2 is the potential
function, since
∂tψ = 2t + u 2 = M, ∂uψ = 2tu = N.
We remark that the Poincaré Lemma does not tell us how to find the potential function, it
only states necessary and sufficient conditions on N and M for the existence of ψ. ⊳
⊳

28 G. NAGY – ODE july 2, 2013
The potential function ψ is crucial to find implicit solutions to exact ODE, as it can be
seen in the following result.
Theorem 1.4.3 (Exact equation). If the differential equation
N(t, y(t)) y ′ (t) + M(t, y(t)) = 0 (1.4.3)
is exact on R = (t1, t2) × (u1, u2), then every solution y must satisfy the algebraic equation
ψ(t, y(t)) = c, (1.4.4)
where c ∈ R and ψ : R → R is a potential function for Eq. (1.4.3).
Proof of Theorem 1.4.3: Since the differential equation in (1.4.3) is exact, Lemma 1.4.2
implies that there exists a potential function ψ satisfying Eqs. (1.4.2). Hence, we can replace
in the differential equation the functions N and M by ∂yψ and ∂tψ, respectively. That is,
0 = N(t, y(t)) y ′ (t) + M(t, y(t))
= ∂yψ(t, y(t)) d
dt y(t) + ∂tψ(t, y(t))
= d
ψ(t, y(t))
dt
⇔ ψ(t, y(t)) = c,
where c is an arbitrary constant. This establishes the Theorem.
Example 1.4.5: Find all solutions y to the differential equation
2ty(t) y ′ (t) + 2t + y 2 (t) = 0.
Solution: The first step is to verify whether the differential equation is exact. We know
the answer, the equation is exact, we did this calculation before in Example 1.4.1, but we
reproduce it here anyway.
N(t, u) = 2tu ⇒ ∂tN(t, u) = 2u,
M(t, u) = 2t + u 2
⇒ ∂tN(t, u) = ∂uM(t, u).
⇒ ∂uM(t, u) = 2u.
Since the equation is exact, Lemma 1.4.2 implies that there exists a potential function ψ
satisfying the equations
∂uψ(t, u) = N(t, u), (1.4.5)
∂tψ(t, u) = M(t, u). (1.4.6)
We now proceed to compute the function ψ. Integrate Eq. (1.4.5) in the variable u keeping
the variable t constant,
∂uψ(t, u) = 2tu ⇒ ψ(t, u) = 2tu du + g(t),
where g is a constant of integration on the variable u, so g can only depend on t. We obtain
ψ(t, u) = tu 2 + g(t). (1.4.7)
Introduce into Eq. (1.4.6) the expression for the function ψ in Eq. (1.4.7) above, that is,
u 2 + g ′ (t) = ∂tψ(t, u) = M(t, u) = 2t + u 2 ,
g ′ (t) = 2t
Integrate in t the last equation above, and choose the integration constant to be zero,
g(t) = t 2 .
We have found that a potential function is given by
ψ(t, u) = tu 2 + t 2 .

G. NAGY – ODE July 2, 2013 29
Therefore, Theorem 1.4.3 implies that all solutions y satisfy the implicit equation
ty 2 (t) + t 2 = c0,
where c0 ∈ R is an arbitrary constant. Notice that choosing g(t) = t 2 + c has the only effect
of modifying the constant c 0 above.
⊳
Example 1.4.6: Find all solutions y to the equation
sin(t) + t 2 e y(t) − 1 y ′ (t) + y(t) cos(t) + 2te y(t) = 0.
Solution: The first step is to verify whether the differential equation is exact. We know
the answer, the equation is exact, we did this calculation before in Example 1.4.2, but we
reproduce it here anyway.
N(t, u) = sin(t) + t 2 e u − 1 ⇒ ∂tN(t, u) = cos(t) + 2te u ,
M(t, u) = u cos(t) + 2te u
⇒ ∂uM(t, u) = cos(t) + 2te u .
Therefore, the differential equation is exact. Then, Lemma 1.4.2 implies that there exists a
potential function ψ satisfying the equations
∂uψ(t, u) = N(t, u), (1.4.8)
∂tψ(t, u) = M(t, u). (1.4.9)
We know proceed to compute the function ψ. We first integrate in the variable u the
equation ∂uψ = N keeping the variable t constant,
∂uψ(t, u) = sin(t) + t 2 e u − 1 ⇒
sin(t) 2 u
ψ(t, u) = + t e − 1 du + g(t)
where g is a constant of integration on the variable u, so g can only depend on t. We obtain
ψ(t, u) = u sin(t) + t 2 e u − u + g(t).
Now introduce the expression above for the potential function ψ in Eq. (1.4.9), that is,
u cos(t) + 2te u + g ′ (t) = ∂tψ(t, u) = M(t, u) = u cos(t) + 2te u ,
g ′ (t) = 0.
The solution is g(t) = c, with c a constant, but we can always choose that constant to be
zero, so we conclude that
We have obtained the potential function
g(t) = 0.
ψ(t, u) = u sin(t) + t 2 e u − u.
Therefore, Theorem 1.4.3 implies that the solution y satisfies the implicit equation
y(t) sin(t) + t 2 e y(t) − y(t) = c0.
The solution y above cannot be written in explicit form. Notice that choosing g(t) = c has
the only effect of modifying the constant c 0 above. ⊳

30 G. NAGY – ODE july 2, 2013
1.4.3. The integrating factor method. There exist non-exact differential equations that
can be converted into an exact differential equation by multiplying the original equation
by an appropriate function, called integrating factor. One example is the case of linear
differential equations. We have seen in Example 1.4.3 that linear equations with coefficient
a = 0 are not exact. Nevertheless, in Section 1.2 we have obtained solutions to linear
equations multiplying the equation by an appropriate function, called the integrating factor.
This functions converts the original differential equation into a total derivative. Using the
terminology of this Section, the integrating factor transforms a linear equation into an exact
equation. Now we generalize this idea to non-linear differential equations.
Theorem 1.4.4 (Integrating factor). Assume that the differential equation
N(t, y(t)) y ′ (t) + M(t, y(t)) = 0 (1.4.10)
is not exact in the sense that the continuously differentiable functions M, N : R → R satisfy
∂tN(t, u) = ∂uM(t, u) on R = (t1, t2) × (u1, u2). If N = 0 on R and the function
1
∂uM(t, u) − ∂tN(t, u)
N(t, u)
does not depend on the variable u, then the equation
(1.4.11)
µ(t)N(t, y(t)) y ′ (t) + µ(t)M(t, y(t)) = 0 (1.4.12)
is exact, where the function µ : (t1, t2) → R is the solution of the equation
µ ′ (t)
µ(t) =
1
∂uM(t, u) − ∂tN(t, u)
N(t, u)
.
Proof of Theorem 1.4.4: Since the original differential equation in (1.4.10) is not exact,
we multiply this equation by a non-zero function µ,
µ(t)N(t, y) y ′ + µ(t)M(t, y) = 0. (1.4.13)
Introducing the functions Ñ(t, u) = µ(t)N(t, u) and ˜ M(t, u) = µ(t)M(t, u), we see that this
new equation is exact iff
The calculation
∂t Ñ(t, u) = ∂u ˜ M(t, u).
∂t Ñ(t, u) = µ′ (t)N(t, u) + µ(t)∂tN(t, u), ∂u ˜ M(t, u) = µ(t)∂uM(t, u),
says that Eq.(1.4.13) is exact iff
which is equivalent to
and in turns it is equivalent to
µ ′ (t)N(t, u) + µ(t)∂tN(t, u) = µ(t)∂uM(t, u),
µ ′ (t)N(t, u) = µ(t) ∂uM(t, u) − ∂tN(t, u) ,
µ ′ (t)
µ(t) =
1
∂uM(t, u) − ∂tN(t, u)
N(t, u)
.
Since the left-hand side does not depend on the variable u, the equation above is solvable
iff the right-hand side does not depend on this variable u. This observation establishes the
Theorem.

G. NAGY – ODE July 2, 2013 31
Example 1.4.7: Find all solutions y to the differential equation
t 2 + t y(t) y ′ (t) + 3t y(t) + y 2 (t) = 0. (1.4.14)
Solution: We first verify whether this equation is exact:
N(t, u) = t 2 + tu ⇒ ∂tN(t, u) = 2t + u,
M(t, u) = 3tu + u 2
⇒ ∂uM(t, u) = 3t + 2u,
therefore, the differential equation is not exact. We now verify whether the extra condition
in Theorem 1.4.4 holds, that is, whether the function in (1.4.11) is u-independent:
1
∂uM(t, u) − ∂tN(t, u)
N(t, u)
1
=
(t2
(3t + 2u) − (2t + u)
+ tu)
1
= (t + u)
t(t + u)
= 1
t .
We conclude that the function (∂uM −∂tN)/N does not depend on the variable u. Therefore,
Theorem 1.4.4 implies that the differential equation in (1.4.14) can be transformed into an
exact equation multiplying it by the function µ solution of the equation
µ ′ (t) 1
= ∂uM − ∂tN
µ(t) N
= 1
⇒ ln(µ(t)) = ln(t) ⇒ µ(t) = t,
t
where we have chosen in second equation the integration constant to be zero. Then, multiplying
the original differential equation in (1.4.14) by the integrating factor µ we obtain
3t 2 y(t) + t y 2 (t) + t 3 + t 2 y(t) y ′ (t) = 0. (1.4.15)
This latter equation is exact, since
Ñ(t, u) = t 3 + t 2 u ⇒ ∂t Ñ(t, u) = 3t2 + 2tu,
˜M(t, u) = 3t 2 u + tu 2
⇒ ∂u ˜ M(t, u) = 3t 2 + 2tu,
that is, ∂t Ñ = ∂u ˜ M. This establishes that the new equation in (1.4.15) is exact. Therefore,
the solution can be found as we did in the previous examples in this Section. That is, we
find the potential function ψ solution of the equations
∂uψ(t, u) = Ñ(t, u), (1.4.16)
∂tψ(t, u) = ˜ M(t, u). (1.4.17)
From the first equation above we obtain
∂uψ = t 3 + t 2
t 3 2
u ⇒ ψ(t, u) = + t u du + g(t).
Integrating on the right hand side above we arrive to
ψ(t, u) = t 3 u + 1
2 t2 u 2 + g(t).
Introduce the expression above for ψ in Eq. (1.4.17), that is,
3t 2 u + tu 2 + g ′ (t) = ∂tψ(t, u) = ˜ M(t, u) = 3t 2 u + tu 2 ,
g ′ (t) = 0.
A solution to this last equation is g(t) = 0. We conclude that a potential function is
ψ(t, u) = t 3 u + 1
2 t2 u 2 .

32 G. NAGY – ODE july 2, 2013
All solutions y to the differential equation in (1.4.14) satisfy the equation
t 3 y(t) + 1
2 t2 y(t) 2 = c0,
where c0 ∈ R is arbitrary. ⊳
We have seen in Example 1.4.3 that linear differential equations with a = 0 are not
exact. We have seen in Section 1.2 that the integrating factor method transforms the linear
differential equation into a total derivative. Such calculation is now a particular case of
Theorem 1.4.4, as it can be seen in our last Example.
Example 1.4.8: Use Theorem 1.4.4 to find all solutions to the linear differential equation
Solution: We first write the differential equation as
y ′ (t) = a(t) y(t) + b(t), a(t) = 0. (1.4.18)
y ′ + −a(t) y − b(t) = 0.
Now it is simple to verify whether the equation is exact or not. We have seen in Example 1.4.3
that this equation is not exact, since
N(t, u) = 1 ⇒ ∂tN(t, u) = 0,
M(t, u) = −a(t)u − b(t) ⇒ ∂uM(t, u) = −a(t).
However, the function
1
∂uM(t, u) − ∂tN(t, u)
N(t, u)
= −a(t)
is independent of the variable u. Therefore, we can compute the integrating factor µ solving
the equation
µ ′ (t)
µ(t) = −a(t) ⇒ µ(t) = e−A(t)
, A(t) = a(t)dt.
Therefore, the equation
is exact, since
Ñ(t, u) = e −A(t)
˜M(t, u) = −a(t) e −A(t) u − b(t) e −A(t)
e −A(t) y ′ − a(t) e −A(t) y − b(t) e −A(t) = 0 (1.4.19)
⇒ ∂t Ñ(t, u) = −a(t) e−A(t) ,
⇒ ∂u ˜ M(t, u) = −a(t) e −A(t) .
This establishes that the new equation in (1.4.19) is exact. Therefore, the solution can be
found as we did in the previous examples in this Section. That is, we find the potential
function ψ solution of the equations
∂uψ(t, u) = Ñ(t, u), (1.4.20)
∂tψ(t, u) = ˜ M(t, u). (1.4.21)
From the first equation above we obtain
∂uψ = e −A(t)
⇒ ψ(t, u) = e −A(t) du + g(t).
The integrating is trivial, since e −A(t) is u-independent, so we arrive to
ψ(t, u) = e −A(t) u + g(t).

Introduce the expression above for ψ in Eq. (1.4.17), that is,
G. NAGY – ODE July 2, 2013 33
−a(t) e −A(t) u + g ′ (t) = ∂tψ(t, u) = ˜ M(t, u) = −a(t) e −A(t) u − b(t) e −A(t) ,
A function g is then given by
g ′ (t) = −b(t) e −A(t) .
g(t) = −
b(t) e −A(t) dt.
We conclude that a potential function is
ψ(t, u) = e −A(t)
u − b(t) e −A(t) dt.
All solutions y to the differential equation in (1.4.18) satisfy the equation
e −A(t)
y(t) − b(t) e −A(t) dt = c0, where c0 ∈ R is arbitrary. This is the implicit form of the solution. The explicit form is
simple to find, and is given by
y(t) = e A(t)
c0 + b(t) e −A(t)
dt .
This expression agrees with the one in Theorem 1.2.4, when we studied linear equations. ⊳

34 G. NAGY – ODE july 2, 2013
1.4.4. Exercises.
1.4.1.- Consider the equation
(1 + t 2 ) y ′ = −2t y.
(a) Determine whether the differential
equation is exact.
(b) Find every solution of the equation
above.
1.4.2.- Consider the equation
t cos(y) y ′ − 2y y ′ = −t − sin(y).
(a) Determine whether the differential
equation is exact.
(b) Find every solution of the equation
above.
1.4.3.- Consider the equation
y ′ =
−2 − y ety
.
−2y + t ety (a) Determine whether the differential
equation is exact.
(b) Find every solution of the equation
above.
1.4.4.- Consider the equation
(6x 5 − xy) + (−x 2 + xy 2 )y ′ = 0,
with initial condition y(0) = 1.
(a) Find the integrating factor µ that
converts the equation above into an
exact equation.
(b) Find an implicit expression for the
solution y of the IVP.
1.4.5.- Consider the equation
2x 2 y + y
x 2
y ′ + 4xy 2 = 0,
with initial condition y(0) = −2.
(a) Find the integrating factor µ that
converts the equation above into an
exact equation.
(b) Find an implicit expression for the
solution y of the IVP.

G. NAGY – ODE July 2, 2013 35
1.5. Applications
Ordinary differential equations can be used to describe situations that change in time.
For example the radioactive decay of a radioactive substance. The amount of the radioactive
substance present in a sample decays exponentially with time. This function is the solution
of a particularly simple differential equation. The same type of differential equations also
describes how materials cool down. This is the Newton cooling law. The first half of this
Section is dedicated to study these situations. In the second half we study the salt concentration
inside a water tank in the case that salty water is allowed in and out the tank. We
will see that the differential equation that describes the salt in the tank is more complicated
than the equation that appears in the exponential decay of a radioactive substance.
1.5.1. Radioactive decay. A radioactive decay is a process that happens in the nucleus
of certain substances, like Uranium-234, Radium-226, Radon-222, Polonium-218, Lead-214,
Cobalt-60, Carbon-14, etc. The nucleus loses energy by emitting different types of charged
particles. Therefore, there are many types of exponential decays. A nucleus can emit an
alpha particle (Helium nucleus), can emit a proton (Hydrogen nucleus), can emit an electron,
gamma-rays, neutrinos, etc. The radioactive decay of a single atom is impossible to predict,
but the decay of a large number of atoms is predictable. It can be seen in the laboratory that
the rate of change in the amount of a radioactive substance is proportional to the negative
of the amount of substance in the sample. If we denote by N(t) the amount of a radioactive
substance in a sample as function of time, then
N ′ (t) = −a N(t),
where a > 0 is called the decay constant and characterizes the radioactive material. The
differential equation above is simple to solve using the integration factor method. Indeed,
N ′ (t) + a N(t) e at = 0 ⇔ e at N(t) ′ = 0 ⇔ N(t) = N0 e −at ,
where N0 = N(0) is the initial amount of the substance. Radioactive substances are often
characterized not by their decay constant a but by their half-life. The half-life is a time τ
needed for half of the radioactive substance to decay, that is,
N(τ) = N0
2 .
The equation above provides a simple relation between the decay constant a and the half-life
τ, namely,
N0
2 = N0 e −aτ
1
⇒ −aτ = ln ⇒ a =
2
ln(2)
.
τ
Therefore, the solution N can be expressed in terms of the half-life as
N(t) = N0 e (−t/τ) ln(2) ⇒ N(t) = N0 e ln[2(−t/τ) ] ⇒ N(t) = N0 2 −t/τ .
From this last expression is clear that for t = τ we get N(τ) = N0/2.
Example 1.5.1: The Carbon-14 is a radioactive isotope of Carbon that is constantly renovated
in the atmosphere and is accumulated by living organisms. While the organism
lives, the amount of Carbon-14 is held constant, because the decay is compensated with
new amounts being incorporated into the organism. When the organism dies, the amount
of Carbon-14 in its remains decays. The half-life of Carbon-14 can be measured in the
laboratory and is τ = 5730 years. If certain remains are found containing an amount of 14%
of the original amount of Carbon-14, find the date of the remains.

36 G. NAGY – ODE july 2, 2013
Solution: Suppose that t = 0 is set at the time that the organism dies. If the remains
contain 14% of the original amount at the present time t, that means
N(t)
N0
= 14
100 .
Since the amount of Carbon-14 only decays after the organism dies, the amount of Carbon-14
as function of time is given by
N(t) = N0 2 −t/τ ,
where τ = 5730 years is the Carbon-14 half-life. Therefore,
2 −t/τ = 14
100
⇒
t
−
τ = log2(14/100) ⇒ t = τ log2(100/14). We obtain that t = 16, 253 years. The organism died more that 16, 000 years ago. ⊳
1.5.2. Newton’s cooling law. The Newton’s cooling law is an empirical law that describes
how objects cool down. The temperature difference ∆T = T − T0 between the temperature
of an object, T , and the constant temperature of the surrounding medium where it is placed,
Ts, evolves in time t following the equation
(∆T ) ′ = −k(∆T ), T (0) = T0, k > 0.
This is the exponential decay equation for the function ∆T . We know that the solution is
(∆T )(t) = (∆T )0 e −kt , that is,
(T − Ts)(t) = (T0 − Ts) e −kt ⇒ T (t) = (T0 − Ts) e −kt + Ts.
where the constant k depends on the material and the surroundings.
Example 1.5.2: A cup with water at 45 C is placed in the cooler held at 5 C. If after 2
minutes the water temperature is 25 C, when will the water temperature be 15 C? while
Solution: We know that T (t) = (T0 − Ts) e −kt + Ts, and also
T0 = 45, Ts = 5, T (2) = 25.
Find t1 such that T (t1) = 15. First we find k,
Now recall that T (2) = 25 C, that is,
T (t) = (45 − 5) e −kt + 5 ⇒ T (t) = 40 e −kt + 5.
20 = T (2) = 40 e −2k ⇒ ln(1/2) = −2k ⇒ k = 1
2 ln(2).
Having the constant k it is simple to find the time t1 such that T (t1) = 15 C. Indeed,
T (t) = 40 e −t ln(√ 2) + 5 ⇒ 10 = 40 e −t1 ln( √ 2) ⇒ t1 = 4.
1.5.3. Salt in a water tank. Consider the situation described in Fig. 2. The tank contains
a volume V (t) of water at a time t with a salt mass Q(t) dissolved in it. Water is pouring
into the tank at a rate ri(t) having salt concentration qi(t). Water is also leaving the tank
at a rate ro(t) with salt concentration qo(t). We recall that a rate r represents water volume
per unit time, while a concentration q represent salt mass per unit volume. We finally
assume that there is a mixing mechanism in the tank such that the salt entering the tank
is instantaneously mixed. This mixing mechanism then implies that the salt concentration
in the tank is homogeneous at every time.
⊳

water
salt
pipe
V (t)
Q (t)
G. NAGY – ODE July 2, 2013 37
r i
q (t)
i
instantaneously mixed
tank
Figure 2. Description of the water tank problem.
r
o
q (t)
Before stating the problems we are interested in solving, we review the physical units of
the main fields involved in the problem. Denoting by [ri] the units of the quantity ri, then
we have
[ri] = [ro] = Volume
Time , [qi] = [qo] = Mass
, [V ] = Volume, [Q] = Mass.
Volume
1.5.4. The Main equations. The mass conservation law provides the main equations of
the mathematical description of the experimental situation above. The main equations are
the following,
V ′ (t) = ri(t) − ro(t), (1.5.1)
Q ′ (t) = ri(t) qi(t) − ro(t), qo(t), (1.5.2)
qo(t) = Q(t)
, (1.5.3)
V (t)
r ′ i(t) = r ′ o(t) = 0. (1.5.4)
The first and second equations above are just the mass conservation of water and salt,
respectively. Water volume and mass are proportional, so both are conserved, and we
chose the volume to write down this conservation in Eq. (1.5.1). This equation is indeed
a conservation because it says that the water volume variation in time is equal to the
difference of volume time rates coming in and going out of the tank. Eq. (1.5.2) is the salt
mass conservation, since the salt mass variation in time is equal to the difference of the
salt mass time rates coming in and going out of the tank. The product of a rate r times
a concentration q has units of mass per time and represents the amount of salt entering or
leaving the tank per unit time. Eq.(1.5.3) is implied by the instantaneous mixing mechanism
in the tank. Since the salt is mixed instantaneously in the tank, the salt concentration in
the tank is homogeneous with value Q(t)/V (t). Finally the equations in (1.5.4) say that
both rates in and out are time independent, that is, constants.
We now study the solutions of Eqs. (1.5.1)-(1.5.4). Since Eq. (1.5.4) says that the water
rates are constant, we denote them as ri and ro. This information in Eq. (1.5.1) implies
that V ′ (t) is constant, so we can easily integrate this equation to obtain
V (t) = (ri − ro) t + V0, (1.5.5)
o

38 G. NAGY – ODE july 2, 2013
where V0 = V (0) is the water volume in the tank at the initial time t = 0. On the other
hand, Eqs.(1.5.2) and (1.5.3) imply that
Q ′ (t) = ri qi(t) − ro
V (t) Q(t).
Since V (t) is known from Eq. (1.5.5), we conclude that the function Q must be solution of
the differential equation
Q ′ (t) = ri qi(t) −
Q(t).
(ri − ro) t + V0
This is a linear ODE for the function Q. Indeed, introducing the functions
the differential equation for Q has the form
ro
ro
a(t) =
, b(t) = ri qi(t), (1.5.6)
(ri − ro) t + V0
Q ′ (t) = −a(t) Q(t) + b(t). (1.5.7)
Solutions of this equation can be obtained with the integrating factor method.
1.5.5. Predictions in particular cases. Instead of finding the solution Q of Eq. (1.5.7)
for the most general functions a and b given in Eq. (1.5.6), we find solutions in particular
cases given by specific choices of the rate constants ri, ro, the concentration function qi, and
the initial data constants V0 and Q0 = Q(0). The study of solutions to Eq. (1.5.7) in several
particular cases might provide a deeper understanding of the physical situation under study
than the expression of the solution Q in the most general case.
Example 1.5.3: Consider the particular case given by the rates ri = ro = r, the incoming
concentration function qi is constant, and the initial water volume in the tank V0 is given.
Then, find the solution to the IVP
Q ′ (t) = −a(t) Q(t) + b(t), Q(0) = Q0,
where function a and b are given in Eq. (1.5.6). Graph the solution function Q for different
values of the initial condition Q0.
Solution: The assumption ri = ro = r implies that the function a is constant, while the
assumption that qi is constant implies that the function b is also constant,
ro
a(t) =
(ri − ro) t + V0
⇒ a(t) = r
V0
= a0,
b(t) = ri qi(t) ⇒ b(t) = ri qi = b0.
We then must solve the IVP for the following constant coefficients linear ODE,
Q ′ (t) = −a0 Q(t) + b0, Q(0) = Q0,
The solution is found using the integrating factor method, the result was found in Theorem
1.1.5 and is given by
Q(t) = Q0 − b0
e −a0t + b0
.
a0
In our case the we can evaluate the constant b0/a0, and the result is
b0
a0
V0
= (rqi)
r
⇒ b0
a0
= qiV0.
Then, the solution Q has the form,
Q(t) = −rt/V0
Q0 − qiV0 e + qiV0. (1.5.8)
a0

G. NAGY – ODE July 2, 2013 39
The initial amount of salt Q0 in the tank can be any non-negative real number. The solution
behaves differently for different values of Q0. We classify these values in three classes:
(a) The initial amount of salt in the tank has the critical value Q0 = qiV0. In this case the
solution Q remains constant equal to this critical value, that is, Q(t) = qiV0.
(b) The initial amount of salt in the tank is bigger than the critical value, that is, Q0 > qiV0.
In this case the salt in the tank Q(t) decreases exponentially towards the critical value.
(c) The initial amount of salt in the tank is smaller than the critical value, that is, Q0 < qiV0.
In this case the salt in the tank Q(t) increases exponentially towards the critical value.
The graphs of few solutions in these three classes are plotted in Fig. 3. ⊳
Q
Q
Q
Q
Q
0
0
Q = q v
0 i 0
0
0
Figure 3. The graph of the function q given in Eq. (1.5.8) for different
values of the initial condition Q0.
Example 1.5.4: Consider the particular case where the rates ri = ro = r, fresh water is
coming into the tank, that is, qi = 0. Then, find the time t1 such that the salt concentration
in the tank Q(t)/V (t) is 1% the initial value.
Solution: The first step to find the time t1 is to find the solution Q(t) to the IVP
Q ′ (t) = −a(t) Q(t) + b(t), Q(0) = Q0,
where function a and b are given in Eq. (1.5.6). In this case we have
ro
a(t) =
(ri − ro) t + V0
⇒ a(t) = r
,
V0
b(t) = ri qi(t) ⇒ b(t) = 0.
The IVP we need to solve is
The solution is given by
Q ′ (t) = − r
V0
Q(t), Q(0) = Q0.
Q(t) = Q0 e −rt/V0 .
We can now proceed to find the time t1. We first need to find the concentration Q(t)/V (t).
This is simple to do, since we already have Q(t), and V (t) = V0, since ri = ro. Therefore,
The condition that defines t1 is
Q(t) Q(t)
= =
V (t) V0
Q0
e
V0
−rt/V0 .
Q(t1) 1
=
V (t1) 100
Q0
V0
.
t

40 G. NAGY – ODE july 2, 2013
From these two equations above we conclude that
1 Q0
=
100 V0
Q(t1) Q0
= e
V (t1) V0
−rt1/V0 .
The time t1 is simple to obtain, since
1
100 = e−rt1/V0 ⇔
1
ln
100
The final result is given by
= − rt1
V0
t1 = V0
r ln(100).
⇔ ln(100) = rt1
.
Example 1.5.5: Consider the particular case where the rates ri = ro = r; there is initially
fresh water in the tank, that is, Q0 = 0; the initial water volume in the tank V0 is given;
and the incoming salt concentration is the function
Then, find the salt in the tank Q(t).
qi(t) = 2 + sin(2t).
Solution: We need to find the solution Q(t) to the IVP
Q ′ (t) = −a(t) Q(t) + b(t), Q(0) = 0,
where function a and b are given in Eq. (1.5.6). In this case we have
ro
a(t) =
(ri − ro) t + V0
⇒ a(t) = r
b(t) = ri qi(t) ⇒
= a0,
V0
b(t) = r 2 + sin(2t) .
The IVP we need to solve is
Q ′ (t) = −a0 Q(t) + b(t), Q(0) = 0.
The solution is computed using the integrating factor method and the result is
Q(t) = e −a0t
t
0
e a0s b(s) ds,
where we already introduced the initial condition Q0 = 0. Introducing the form of the
function b we obtain the expression
Q(t) = e −a0t
t
0
e a0s 2 + sin(2s) ds.
This is the solution of the problem. We only need to compute the integral given in the
equation above. This is not a difficult calculation, although it is not straightforward. We
start with the following integral found in an integration table,
0
e ks sin(ls) ds = eks
k2 + l2
k sin(ls) − l cos(ls) ,
where k and l are constants. Therefore,
t
e a0s 2 + sin(2s)
2
ds = e a0s t a0s
e
+
a0
0
a2 0 + 22
= 2 a0t e
e − 1 +
a0
a0t
a2 0 + 22
V0
a0 sin(2s) − 2 cos(2s) t
,
a0 sin(2t) − 2 cos(2t) +
0
a 2 0
2
.
+ 22
⊳

G. NAGY – ODE July 2, 2013 41
With the integral above we can compute the solution Q as follows,
Q(t) = e −a0t 2 a0t
e − 1 +
ea0t
a0 sin(2t) − 2 cos(2t) +
a0
a2 0 + 22
recalling that a0 = r/V0. We rewrite expression above as follows,
Q(t) = 2 1
+ a0 sin(2t) − 2 cos(2t)
+ 22
2 2
+ −
+ 22
a0
a 2 0
y
1
Q ( t )
− 2 t
1 − ( 3 / 4 ) e
a 2 0
t
a0
2
a2 0 + 22
,
e −a0t . (1.5.9)
Figure 4. The graph of the function Q given in Eq. (1.5.9) for a0 = 2.
⊳

42 G. NAGY – ODE july 2, 2013
1.5.6. Exercises.
1.5.1.- A radioactive material decays at
a rate proportional to the amount
present. Initially there is 50 milligrams
of the material present and after one
hour the material has lost 80% of its
original mass.
(a) Find the mass of the material as
function of time.
(b) Find the mass of the material after
four hours.
(c) Find the half-life of the material.
1.5.2.- A tank initially contains V0 = 100
liters of water with Q0 = 25 grams of
salt. The tank is rinsed with fresh water
flowing in at a rate of ri = 5 liters
per minute and leaving the tank at the
same rate. The water in the tank is wellstirred.
Find the time such that the
amount the salt in the tank is Q1 = 5
grams.
1.5.3.- A tank initially contains V0 = 100
liters of pure water. Water enters the
tank at a rate of ri = 2 liters per minute
with a salt concentration of q1 = 3
grams per liter. The instantaneously
mixed mixture leaves the tank at the
same rate it enters the tank. Find the
salt concentration in the tank at any
time t 0. Also find the limiting
amount of salt in the tank in the limit
t → ∞.
1.5.4.- A tank with a capacity of Vm = 500
liters originally contains V0 = 200 liters
of water with Q0 = 100 grams of salt
in solution. Water containing salt with
concentration of qi = 1 gram per liter
is poured in at a rate of ri = 3 liters
per minute. The well-stirred water is
allowed to pour out the tank at a rate
of ro = 2 liter per minute. Find the
salt concentration in the tank at the
time when the tank is about to overflow.
Compare this concentration with the
limiting concentration at infinity time
if the tank had infinity capacity.

G. NAGY – ODE July 2, 2013 43
1.6. Non-linear equations
This short Section intends to highlight the main differences between solutions to linear
differential equations and solutions to non-linear differential equations. The former equations
were studied in Section 1.2 and several examples of non-linear equations were studied
in Sections 1.3 and 1.4. The plan here is first, to review the main result on solutions to
an initial value problem involving a linear ODE, already given in Theorem 1.2.4. From
this result we highlight the main properties shared by solutions of linear ODE, listed in the
items (a)-(c) below. Then we define what is a non-linear ODE and we state that none of the
properties in (a)-(c) below hold for solutions of non-linear ODE. We prove this statement
showing three examples of non-linear ODE with solutions having the properties listed in
items (i)-(iii) below, respectively. We finally state in Theorem 1.6.2 below, without proof,
the main result about existence and uniqueness of solutions to an initial value problem
involving a non-linear ODE.
1.6.1. On solutions of linear differential equations. The initial value problem for a
linear differential equation is the following: Given continuous functions a, b : (t1, t2) → R,
with t2 > t1, and constants t0 ∈ (t1, t2), y0 ∈ R, find the function y : (t1, t2) → R solution
of the equations
y ′ = −a(t) y + b(t), y(t0) = y0. (1.6.1)
The main result regarding solutions to this problem is summarized in Theorem 1.2.4, which
we reproduce it below.
Theorem 1.2.4 (Variable coefficients). Given continuous functions a, b : (t1, t2) → R
and constants t 0 ∈ (t 1, t 2) and y 0 ∈ R, the initial value problem
has the unique solution y : (t1, t2) → R given by
y ′ = a(t) y + b(t), y(t0) = y0, (1.2.6)
y(t) = e A(t)
y0 +
where we introduced the function A(t) =
t
t0
t
t0
a(s) ds.
e −A(s)
b(s) ds , (1.2.7)
We highlight now three main properties that solutions to a linear initial value problem
have according to Theorem 1.2.4.
Properties of solutions to linear differential equations:
(a) There is an explicit expression for the solutions of a linear IVP, given in Eq. (1.2.7).
(b) For every initial condition y 0 ∈ R there exists a unique solution to a linear IVP.
(c) For every initial condition y 0 ∈ R the corresponding solution y(t) of a linear IVP is
defined for all t ∈ (t1, t2).
None of these properties hold for solutions of non-linear differential equations. We will
see that there exist non-linear differential equations having solutions that cannot be written
in explicit form. We will also see that an initial value problem for a non-linear differential
equation has two solutions instead of a unique solution. We will finally see that the solution
to an initial value problem for a non-linear equation is defined in a domain that depends on
the initial data.

44 G. NAGY – ODE july 2, 2013
1.6.2. On solutions of non-linear differential equations. We now state more precisely
what is a non-linear ordinary differential equation and we then present the main result about
the solutions of an IVP for a non-linear ODE.
Definition 1.6.1. An ordinary differential equation y ′ (t) = f(t, y(t)) is called non-linear
iff the function (t, u) ↦→ f(t, u) is non-linear in the second argument.
Example 1.6.1:
(a) The differential equation
y ′ (t) = t2
y 3 (t)
is non-linear, since the function f(t, u) = t 2 /u 3 is non-linear in the second argument.
(b) The differential equation
y ′ (t) = 2ty(t) + ln y(t)
is non-linear, since the function f(t, u) = 2tu + ln(u) is non-linear in the second argument,
due to the term ln(u).
(c) The differential equation
y ′ (t)
= 2t2
y(t)
is linear, since the function f(t, u) = 2t 2 u is linear in the second argument.
The initial value problem involving a non-linear ODE has a unique solution in the case
that the ODE meets certain conditions. The precise statement is summarized below.
Theorem 1.6.2 (Non-linear ODE). Fix a non-empty rectangle R = (t1, t2)×(u1, u2) ⊂ R 2
and fix a function f : R → R denoted as (t, u) ↦→ f(t, u). If the function f and its derivative
∂uf are continuous on R, and the point (t0, y0) ∈ R, then there exists a smaller open
rectangle ˆ R ⊂ R with (t0, y0) ∈ ˆ R such that the IVP
has a unique solution y on the set ˆ R ⊂ R 2 .
y ′ (t) = f(t, y(t)), y(t0) = y0 (1.6.2)
We do not provide the proof of this Theorem, since it requires ideas beyond the level of
these notes. This statement is the generalization of Theorem 1.2.4 from linear to non-linear
ODE. We now highlight three main properties of the solutions to non-linear ODE, properties
that are allowed by the statement in Theorem 1.6.2.
Properties of solutions to non-linear differential equations:
(i) There is no general explicit expression for the solution y(t) to a non-linear differential
equation.
(ii) Non-uniqueness of solution to the IVP in Eq. (1.6.2) may happen at points (t, u) ∈ R 2
where ∂uf is not continuous.
(iii) Changing the initial data y 0 in Eq. (1.6.2) may change the domain on the variable t
where the solution y(t) is defined.
The next three examples (1.6.2)-(1.6.4) below address the statements in (i)-(iii), respectively.
In the first example we have an equation whose solution cannot be written in explicit
form. The reason is not lack of ingenuity, it is actually proven that such explicit expression
does not exist.
⊳

G. NAGY – ODE July 2, 2013 45
Example 1.6.2: Given non-zero constants a1, a2, a3, a4, find every solution y to the ODE
y ′ (t) =
t2
y4 (t) + a4 y3 (t) + a3 y2 . (1.6.3)
(t) + a2 y(t) + a1
Solution: We see that the non-linear differential equation above is separable, so we can
find solutions following the ideas given in Sect. 1.3. That is, we first rewrite the equation as
4
y (t) + a4 y 3 (t) + a3 y 2 ′ 2
(t) + a2 y(t) + a1 y (t) = t ,
then we integrate on both sides of the equation,
y4 (t) + a4 y 3 (t) + a3 y 2
′
(t) + a2 y(t) + a1 y (t) dt =
Introduce the substitution u = y(t), so du = y ′ (t) dt,
(u 4 + a4 u 3 + a3 u 2
+ a2 u + a1 du =
t 2 dt + c.
t 2 dt + c.
Integrate and in the result substitute back the function y, hence we obtain
1
5 y5 (t) + a4
4 y4 (t) + a3
3 y3 (t) + a2
2 y(t) + a1 y(t) = t3
+ c.
3
This is an implicit form for the solution y of the problem, given by the root of a polynomial
degree five. Since there is no formula for the roots of a general polynomial degree bigger or
equal five, we see that there is no explicit expression for solutions y of Eq. (1.6.3). ⊳
We now give an example of the statement in (ii). We consider a differential equation
defined by a function f that does not satisfy one hypothesis in Theorem 1.6.2. The function
values f(t, u) have a discontinuity at a line in the (t, u) plane where the initial condition for
the initial value problem is given. We then show that such initial value problem has two
solutions instead of a unique solution.
Example 1.6.3: Find every solution y of the initial value problem
y ′ (t) = y 1/3 (t), y(0) = 0. (1.6.4)
Remark: The equation above is non-linear and separable, and the function f(t, u) = u1/3 has derivative
∂uf = 1 1
,
3 u2/3 so the function ∂uf is not continuous at u = 0.
Solution: The solution to the initial value problem in Eq. (1.6.4) exists but it is not unique,
since we now show that it has two solutions. The first solution is
y 1(t) = 0.
The second solution can be computed as using the ideas from separable equations, that is,
y(t) −1/3 ′
y (t) dt = dt + c0.
Then, the substitution u = y(t), with du = y ′ (t) dt, implies that
u −1/3
du = dt + c0. Integrate and substitute back the function y. The result is
3
2/3 y(t) = t + c0,
2

46 G. NAGY – ODE july 2, 2013
so the solution is
The initial condition above implies
so the second solution is:
y(t) =
0 = y(0) =
2
3 (t + c 3/2 0) .
2
3 c 3/2 0
y2(t) =
2
3 t
3/2 .
⇒ c 0 = 0,
Finally, an example of the statement in (iii). We show that the solution of an initial value
problem for a non-linear equation is defined in a domain that depends on the initial data.
Example 1.6.4: Find the solution y to the initial value problem
y ′ (t) = y 2 (t), y(0) = y 0.
Solution: This is a non-linear separable equation, so we can again apply the ideas in
Sect. 1.3. Performing all the calculations in one line, we summarize them as follows,
′ y (t) dt
y2 (t) =
dt + c0 ⇒ − 1
y(t) = t + c0 ⇒ y(t) = − 1
c0 + t .
Using the initial condition in equation above we obtain
y0 = y(0) = − 1
⇒ c0 = − 1
.
c0
So the solution of the initial value problem above is:
1
y(t) =
1
.
− t
y0 This solution diverges at t = 1/y0, so the domain of the solution y is not the whole real line
R. Instead, the domain is R − {y0}, so it depends on the values of the initial data y0. ⊳
The last example tries to help understand the continuity hypothesis given in Theorem
1.6.2 on the nonlinear function f that defines the differential equation.
Example 1.6.5: Consider the non-linear IVP:
y ′ 1
(t) =
(t − 1)(t + 1)(y(t) − 2)(y(t) + 3) , y(t0) = y0. (1.6.5)
Find the regions on the plane R 2 where Theorem 1.6.2 does not hold.
Solution: In this case the function f is given by:
1
f(t, u) =
, (1.6.6)
(t − 1)(t + 1)(u − 2)(u + 3)
therefore, f is not defined on the lines
t = 1, t = −1, u = 2, u = −3.
where we denote (t, u) ∈ R 2 . See Fig. 5. For example, in the case that the initial data is
t 0 = 0, y 0 = 1, then Theorem 1.6.2 implies that there exists a unique solution on any region
ˆR contained in the rectangle R = (−1, 1) × (−3, 2). On the other hand, if the initial data
for the IVP in Eq. (1.6.5) is t = 0, y0 = 2, then the hypothesis of Theorem 1.6.2 are not
satisfied. ⊳
y0
⊳

G. NAGY – ODE July 2, 2013 47
t=−1
−1
u
1
−1
(0,2)
(0,1)
R
1
t=1
u=2
u=−3
Figure 5. Regions in the plane where the function f defined in Eq. (1.6.6)
is not defined.
Summary: Both Theorems 1.2.4 and 1.6.2 state that there exist solutions to linear and
non-linear differential equations, respectively. However, Theorem 1.2.4 provides more information
about the solutions to a reduced type of equations, linear problems; while Theorem
1.6.2 provides less information about solutions to wider type of equations, linear and
non-linear. Once again:
• Linear differential equations:
(a) There is an explicit expression for the solution of a linear IVP.
(b) For every initial condition y0 ∈ R there exists a unique solution to a linear IVP.
(c) The domain of the solution of a linear IVP is defined for every initial condition
y0 ∈ R.
• Non-linear differential equations:
(i) There is no general explicit expression for the solution y(t) to a non-linear
differential equation.
(ii) Non-uniqueness of solution to a non-linear IVP may happen at points (t, u) ∈
R 2 where ∂uf is not continuous.
(iii) Changing the initial data y0 may change the domain on the variable t where
the solution y(t) is defined.
1.6.3. Direction Fields. One conclusion of the discusion in this section is that non-linear
differential equations are more difficult to solve that the linear ones. It is important to
develop methods to obtain any type of information from the solution of a differential equation
without having to actually solve the equation. One of such method is based on the idea of
direction fields. We focus this discussion on differential equations of the form
y ′ (t) = f(t, y(t)).
One does not need to solve the differential equation above to have a qualitative idea of the
solution. We only need to recall that y ′ (t) represents the slope of the tangent line to the
graph of function y at the point (t, y(t)) in the yt-plane. Therefore, what the differential
equation above provides is all these slopes, f(t, y(t)), for every point (t, y(t)) in the yt-plane.
The Key idea is this: Graph the function f(t, y) on the yt-plane, not as points, but as slopes
of small segments.
t

48 G. NAGY – ODE july 2, 2013
Definition 1.6.3. A direction field for the differential equation y ′ (t) = f(t, y(t)) is the
graph on the yt-pane of the values f(t, y) as slopes of a small segments.
Example 1.6.6: We know that the solution of y ′ = y are given by the exponentials y(t) =
y0 e t , for ever constant y0 ∈ R. The graph of these solution is simple. So is the direction
field shown in Fig. 6. ⊳
Figure 6. Direction field for the equation y ′ = y.
Example 1.6.7: The solution of y ′ = sin(y) is simple to compute. The equation is separable.
After some calculations the implicit solution are
ln csc(y0) + cot(y)
= t.
csc(y) + cot(y)
for y0 ∈ R. The graph of these solution is not simple to do. But the direction field is simple
to plot and can be seen in Fig. 7. ⊳
Example 1.6.8: The solution of y ′ = (1 + y3 )
(1 + t2 could be hard to compute. But the direction
)
field is simple to plot and can be seen in Fig. 8. ⊳

G. NAGY – ODE July 2, 2013 49
Figure 7. Direction field for the equation y ′ = sin(y).
Figure 8. Direction field for the equation y ′ = (1 + y 3 )/(1 + t 2 ).

50 G. NAGY – ODE july 2, 2013
1.6.4. Exercises.
1.6.1.- By looking at the equation coefficients,
find a domain where the solution
of the initial value problem below exists,
(a) (t 2 −4) y ′ +2 ln(t) y = 3t, and initial
condition y(1) = −2.
(b) y ′ y
= , and initial condition
t(t − 3)
y(−1) = 2.
1.6.2.- State where in the plane with points
(t, y) the hypothesis of Theorem 1.6.2
are not satisfied.
(a) y ′ =
y2
2t − 3y .
(b) y ′ = 1 − t2 − y2 .
1.6.3.- Find the domain where the solution
of the initial value problems below is
well-defined.
(a)
y ′ = −4t
, y(0) = y0 > 0.
y
(b)
y ′ = 2ty 2 , y(0) = y0 > 0.

G. NAGY – ODE July 2, 2013 51
Chapter 2. Second order linear equations
In Chapter 1 we studied methods to find solutions to particular types of first order
differential equations. A first oder differential equation is an equation where besides the
function only its first derivative appear in the equation. In this Chapter we study second
order differential equations, that is, differential equations where the second derivative of the
unknown function also appears in the equation. The whole Chapter is dedicated to study
linear second order equations only. This observation suggests that second order equations
are more complicated to solve than first order equations, since we studied and solved linear
first order equations in just the first Section of Chapter 1.
2.1. Variable coefficients
2.1.1. Main definitions and properties. In this Section we present the precise definition
of linear second order differential equations and few properties of their solutions. The central
result, Theorem 2.1.6, asserts that there exist solutions to this type of equations in the case
that the equation coefficients are continuous functions. We present the statement without
proof, since the proof requires functional analysis techniques beyond the scope of this class.
Definition 2.1.1. Given functions a1, a0, b : (t1, t2) → R, the differential equation in the
unknown function y : (t1, t2) → R given by
y ′′ + a 1(t) y ′ + a 0(t) y = b(t) (2.1.1)
is called a second order linear differential equation. The equation in (2.2.1) is called
homogeneous iff for all t ∈ R holds
b(t) = 0.
The equation in (2.2.1) is called of constant coefficients iff a1, a0, and b are constants,
otherwise the equation is called of variable coefficients.
We remark that the notion of an homogeneous equation presented here is completely
different from the homogeneous equations we studied in Section 1.3.
Example 2.1.1:
(a) A second order, linear, homogeneous, constant coefficients equation is
y ′′ + 5y ′ + 6 = 0.
(b) A second order order, linear, constant coefficients, non-homogeneous equation is
y ′′ − 3y ′ + y = 1.
(c) A second order, linear, non-homogeneous, variable coefficients equation is
y ′′ + 2t y ′ − ln(t) y = e 3t .
(d) Newton’s second law of motion for point particles of mass m moving in one space
dimension under a force f : R → R is given by
m y ′′ (t) = f(t).
This is the equation that says mass times acceleration equals force. The acceleration is
the second time derivative of the position function y. ⊳
Solutions to linear homogeneous equations have the property that any linear combination
of solutions is also a solution. Here is a more precise statement.

52 G. NAGY – ODE july 2, 2013
Theorem 2.1.2 (Superposition). If the functions y 1 and y 2 are solutions to the linear
homogeneous equation
y ′′ + a1(t) y ′ + a0(t) y = 0, (2.1.2)
then the linear combination c1y1(t) + c2y2(t) is also a solution for any constants c1, c2 ∈ R.
Proof of Theorem 2.1.2: Verify that the function y = c 1y 1 + c 2y 2 satisfies Eq. (2.1.2) for
every constants c 1, c 2, that is,
(c1y1 + c2y2) ′′ + a1(t)(c1y1 + c2y2) ′ + a0(t)(c1y1 + c2y2)
= (c 1y ′′
1 + c 2y ′′
2 ) + a 1(t)(c 1y ′ 1 + c 2y ′ 2) + a 0(t)(c 1y 1 + c 2y 2)
′′
= c1 y 1 + a1(t)y ′
1 + a0(t)y1 + c2
y ′′
2 + a 1(t)y ′ 2 + a 0(t)y 2
= 0.
2.1.2. Wronskian and linear dependence. We now introduce the notion of linearly dependent
and linearly independent functions.
Definition 2.1.3. Two continuous functions y1, y2 : (t1, t2) ⊂ R → R are called linearly
dependent on the interval (t1, t2) iff there exists a constant c such that for all t ∈ I holds
y1(t) = c y2(t).
The two functions are called linearly independent on the interval (t1, t2) iff they are not
linearly dependent.
We see that two functions are linearly dependent on an interval iff the functions are
proportional to each other on that interval, and they are linearly independent if they are
not proportional to each other in that interval. Notice that the function zero is proportional
to every function, so any set containing the zero function is linearly dependent.
The definitions of linearly dependent or independent functions found in the literature is
equivalent to the definition given here, but it is worded in a slight different way. One usually
finds in the literature that two functions are called linearly dependent on the interval (t1, t2)
iff there exist constants c1, c2, not both zero, such that for all t ∈ (t1, t2) holds
c 1y 1(t) + c 2y 2(t) = 0.
Two functions are called linearly independent on the interval (t1, t2) iff they are not linearly
dependent, that is, the only constants c1 and c2 that for all t ∈ (t1, t2) satisfy the equation
c 1y 1(t) + c 2y 2(t) = 0
are the constants c1 = c2 = 0. This latter wording makes it simple to generalize these
definitions to an arbitrary number of functions.
Example 2.1.2:
(a) Show that y 1(t) = sin(t), y 2(t) = 2 sin(t) are linearly dependent.
(b) Show that y1(t) = sin(t), y2(t) = t sin(t) are linearly independent.
Solution:
Part (a): This is trivial, since 2y 1(t) − y 2(t) = 0.
Part (b): Find constants c1, c2 such that for all t ∈ R holds
c 1 sin(t) + c 2t sin(t) = 0.
Evaluating at t = π/2 and t = 3π/2 we obtain
c1 + π
2 c2 = 0, c1 + 3π
2 c2 = 0 ⇒ c1 = 0, c2 = 0.
We conclude: The functions y1 and y2 are linearly independent. ⊳

G. NAGY – ODE July 2, 2013 53
We now introduce a function W that determines whether two functions y 1, y 2 are linearly
independent or not. This function W was first introduced by a polish scientist Josef Wronski
in 1821 while studying a different problem, and it is now called the Wronskian.
Definition 2.1.4. The Wronskian of functions y 1, y 2 : (t 1, t 2) → R is the function
Wy1y2 (t) = y1(t)y ′ 2(t) − y ′ 1(t)y2(t).
Remarks:
y1(t) y2(t)
(a) If we introduce the matrix-valued function A(t) =
, then the Wronskian
y ′ 1(t) y ′ 2(t)
can be written using the determinant of a 2 × 2 matrix, that is, Wy1y2 (t) = detA(t) .
An alternative notation is: Wy1y2 =
y1 y2 y ′ 1 y ′
2
.
(b) We will see that the Wronskian determines whether two functions are linearly dependent
or linearly independent.
Example 2.1.3: Find the Wronskian of the functions:
(a) y 1(t) = sin(t) and y 2(t) = 2 sin(t). (ld)
(b) y1(t) = sin(t) and y2(t) = t sin(t). (li)
Solution:
Part (a): By the definition of the Wronskian:
Wy1y2 (t) =
y1(t)
y2(t)
=
sin(t)
2 sin(t)
cos(t) 2 cos(t) = sin(t)2 cos(t) − cos(t)2 sin(t)
y ′ 1(t) y ′ 2(t)
We conclude that Wy1y2 (t) = 0. Notice that y1 and y2 are linearly dependent.
Part (b): Again, by the definition of the Wronskian:
Wy1y2 (t) = sin(t)
t sin(t)
cos(t) sin(t) + t cos(t) = sin(t)sin(t) + t cos(t) − cos(t)t sin(t).
We conclude that Wy1y2 (t) = sin2 (t). Notice that y1 and y2 are linearly independent. ⊳
We can see that the Wronskian detects whether the two functions are linearly dependent
or independent. The Wronskian vanishes identically iff the functions are linearly dependent.
This property is the main result of the following statement.
Lemma 2.1.5 (Wronskian and linearly dependence). The continuously differentiable
functions y 1, y 2 : (t 1, t 2) → R are linearly dependent iff Wy1y2 (t) = 0 holds for all t ∈ (t 1, t 2).
Proof of Lemma 2.1.5:
(⇒) Since the functions y1, y2 form a linearly dependent set, then there exists a nonzero
constant c such that y 1 = cy 2. Therefore,
Wy1y2 (t) = c y 2 y ′ 2 − c y ′ 2 y 2 = 0.
(⇐) Since the Wronskian of the non-zero functions y1, y2 vanishes, we obtain that
0 = Wy1y2 = y1 y ′ 2 − y ′ 1
y2 ⇔ y′ 1
⇔
y1(t)
y2(t)
ln = ln
y1(t0) y2(t0)
y1
= y′ 2
y2
therefore, calling c = y1(t0)/y2(t0), we obtain that
y1(t) = c y2(t).
This establishes the Lemma.

54 G. NAGY – ODE july 2, 2013
Example 2.1.4: Show whether the following two functions form a l.d. or l.i. set:
y 1(t) = cos(2t) − 2 cos 2 (t), y 2(t) = cos(2t) + 2 sin 2 (t).
Solution: In this case it is not clear whether these two functions are proportional to each
other or not. They could be proportional due to a trigonometric identity. Actually this is
the case, as we can see by computing their Wronskian:
Wy1y2 (t) = cos(2t) − 2 cos 2 (t) −2 sin(2t) + 4 sin(t) cos(t)
− − 2 sin(2t) + 4 sin(t) cos(t) cos(2t) + 2 sin 2 (t) ,
but sin(2t) = 2 sin(t) cos(t), which implies that Wy1y2 (t) = 0. We conclude that the functions
y1 and y2 are linearly dependent. ⊳
2.1.3. Fundamental and general solutions. Second order linear differential equations
have solutions in the case that the equation coefficients are continuous functions. We present
the statement without proof.
Theorem 2.1.6 (Variable coefficients). If a 1, a 0, b : (t 1, t 2) → R are continuous, then
there exist two linearly independent solutions y 1, y 2 : (t 1, t 2) → R to the equation
y ′′ + a1(t) y ′ + a0(t) y = b(t). (2.1.3)
Furthermore, for every constant t0 ∈ (t1, t2) and y0, y1 ∈ R, there exists a unique solution
y : (t1, t2) → R to the initial value problem given by Eq. (2.1.3) with the initial conditions
y(t 0) = y 0 and y ′ (t 0) = y 1.
Remarks:
(a) Unlike the first order linear differential equations where we have an explicit expression
for the solution, Theorem 1.2.4, there is no explicit expression for the solution of second
order linear equation in the Theorem above. This suggests that second order equations
are more difficult to solve than first order equations.
(b) We will provide a formula for solutions to second order linear equations though, in
the particular case that the equation has constant coefficients. We study this case in
Section 2.2 and the result is Theorem 2.2.2.
(c) Two integrations must be done to find solutions to second order linear differential equations.
Therefore, initial value problems with two initial conditions can have a unique
solution.
Example 2.1.5: Find the longest interval I ∈ R such that there exists a unique solution to
the initial value problem
(t − 1)y ′′ − 3ty ′ + 4y = t(t − 1), y(−2) = 2, y ′ (−2) = 1.
Solution: We first write the equation above in the form given in Theorem (2.1.6,
y ′′ − 3t
t − 1 y′ + 4
y = t.
t − 1
The intervals where the hypotheses in Theorem 2.1.6 are satisfied, that is, where the equation
coefficients are continuous, are I1 = (−∞, 1) and I2 = (1, ∞). Since the initial condition
belongs to I1, the solution domain is
I 1 = (−∞, 1). ⊳

G. NAGY – ODE July 2, 2013 55
The Theorem above and the Wronskian of two functions are the key ideas needed to
find properties of the solutions to differential equations, in the case where the solutions are
not explicitly known. For example, the only information needed from the solutions in the
statement below is that they are continuous functions.
Theorem 2.1.7 (Linear independent solutions). If the functions a 1, a 0 : (t 1, t 2) → R
are continuous, then each initial value problem
y ′′
1 + a1(t) y ′ 1 + a0(t) y1 = 0, y1(0) = 1, y ′ 1(0) = 0,
y ′′
2 + a1(t) y ′ 2 + a0(t) y2 = 0, y2(0) = 0, y ′ 2(0) = 1,
has a unique solution y1, y2 : (t1, t2) → R for any t0 ∈ (t1, t2), respectively, and these
solutions are linearly independent.
Proof of Theorem 2.1.7: We know that the solutions actually exist because of Theorem
2.1.6. We now only compute the Wronskian at t0, that is,
Wy1y2 (t 0) = y 1(t 0) y ′ 2(t 0) − y ′ 1(t 0) y 2(t 0) = 1.
Since the Wronskian is a continuous function, it will be non-zero in a neighborhood of t0.
Then, Lemma 2.1.5 establishes the Theorem.
The superposition property says that every linear combination y(t) = c1 y1(t) + c2 y2(t),
is also a solution of the differential equation
y ′′ + a1(t) y ′ + a0(t) y = 0,
Conversely, one can prove that every solution y of the equation above can be written as a
linear combination of the solutions y 1, y 2. These comments suggest the following definitions.
Definition 2.1.8. Two linearly independent solutions y1, y2 of the homogeneous equation
y ′′ + a 1(t)y ′ + a 0(t)y = 0,
are called fundamental solutions of the differential equation. Given any two fundamental
solutions y1, y2, and arbitrary constants c1, c2, the function
y(t) = c 1 y 1(t) + c 2 y 2(t)
is called the general solution of the homogeneous differential equation.
Example 2.1.6: Find two linearly independent solutions to the equation
y ′′ + y ′ − 2y = 0.
Solution: One way is to find any two solutions, and later on verify that they are linearly
independent. For example, we look for solutions of the form y(t) = ert . Then, we obtain
the characteristic equation
r 2 + r − 2 = 0 ⇒ r = 1
√ r1 = 1
−1 ± 1 + 8 ⇒
2
r2 = −2.
Therefore, y1(t) = e t , y2(t) = e −2t are linearly independent solutions to the equation above.
Another way to solve this problem is to use Theorem 2.1.7. This Theorem says that two
linearly independent solutions of the equation above cane be obtained by solving the two
initial value problems mentioned in that Theorem. The respective solutions are
y1(t) = 2
3 et + 1
3 e−2t , y2(t) = 1
t −2t
e − e
3
.
⊳

56 G. NAGY – ODE july 2, 2013
2.1.4. The Abel Theorem. Theorem 2.1.7 says that the functions y 1 and y 2 that solve
the initial values problems in that Theorem are linearly independent in a neighborhood of
t0. We extend now this result from a neighborhood of t0 to the whole domain (t1, t2) where
the solution is defined. The extension is based in Abel’s Theorem, which also answers the
second question asked earlier in this Section.
Theorem 2.1.9 (Abel). Let a 1, a 0 : (t 1, t 2) → R be continuous functions, and y 1, y 2 be
twice continuously differentiable solutions of the equation
Then, the Wronskian Wy1y2
y ′′ + a1(t) y ′ + a0(t) y = 0. (2.1.4)
is a solution of the equation
W ′ y1y2 (t) + a1(t) Wy1y2 (t) = 0.
Therefore, for any t0 ∈ (t1, t2), the Wronskian Wy1y2
where W0 = Wy1y2 (t0) and A1(t) =
Wy1y2 (t) = W0 e A1(t) ,
t
t0
a1(s) ds.
is given by the expression
This Theorem says that the Wronskian of two solutions of Eq. (2.1.4) is non-zero at a
single point t0 ∈ (t1, t2) iff it is non-zero on the whole interval (t1, t2) where Wy1y2 is defined.
In other words, if the two solutions y1, y2 are linearly independent in a neighborhood of a
point t0, then they are linearly independent on the whole interval (t1, t2) where they are
defined. This statement answers the second question above.
Example 2.1.7: Find the Wronskian of two solutions of the equation
t 2 y ′′ − t(t + 2) y ′ + (t + 2) y = 0, t > 0.
Solution: Notice that we do not known the explicit expression for the solutions. Nevertheless,
Theorem 2.1.9 says that we can compute their Wronskian. First, we have to rewrite
the differential equation in the form given in that Theorem, namely,
y ′′
2
− + 1 y
t ′
2 1
+ + y = 0.
t2 t
Then, Theorem 2.1.9 says that the Wronskian satisfies the differential equation
W ′
2
y1y2 (t) − + 1 Wy1y2 (t) = 0.
t
This is a first order, linear equation for Wy1y2 , so its solution can be computed using the
method of integrating factors. That is, first compute the integral
t
2
t
− + 1 ds = −2 ln − (t − t0) t0 s t0 2 t
0
= ln − (t − t0).
Then, the integrating factor µ is given by
t 2
µ(t) = t2 0
t 2 e−(t−t0) ,
which satisfies the condition µ(t0) = 1. So the solution, Wy1y2 is given by
µ(t)Wy1y2 (t)
′
= 0 ⇒ µ(t)Wy1y2 (t) − µ(t0)Wy1y2 (t0) = 0

so, the solution is
G. NAGY – ODE July 2, 2013 57
Wy1y2
t2
(t) = Wy1y2 (t0) e (t−t0)
.
If we call the constant c = Wy1y2 (t0)/[t 2 0e t0], then the Wronskian has the simpler form
t 2 0
Wy1y2 (t) = c t2 e t .
Proof of Theorem 2.1.9: Recall that both y1 and y2 are solutions to Eq. (2.1.4), that is
y ′′
1 + a1(t) y ′ 1 + a0(t) y1 = 0
y ′′
2 + a1(t) y ′ 2 + a0(t) y2 = 0.
Now, multiply the first equation above by −y2, the second equation by y1, and then add the
resulting equations together. One obtains
Since
(y1 y ′′
2 − y ′′
1 y2) + a1(t) (y1 y ′ 2 − y ′ 1 y2) = 0.
Wy1y2 = y1 y ′ 2 − y ′ 1 y2 ⇒ W ′ y1y2 = y1 y ′′
2 − y ′′
1 y2,
we obtain the equation
W ′ y1y2 + a1(t) Wy1y2 = 0.
This is a first order linear equation, and its solution can be found using the method of
integrating factors. The result is the expression given in the Theorem. This establishes the
Theorem.
2.1.5. Special Second order nonlinear equations. Just for completeness, we now take
the time to introduce definiton of nonlinear second order differential equations.
Definition 2.1.10. Given a functions f : R 3 → R, a second order differential equation in
the unknown function y : R → R is given by
y ′′ = f(t, y, y ′ ).
The equation is linear iff function f is linear in the arguments y and y ′ .
One could say that nonlinear second order differential equation are usually more difficult
to solve than first order nonlinear equations. However, there are two particular cases where
second order equations can be transformed into first order equations.
(a) y ′′ = f(t, y ′ ). The function y is missing.
(b) y ′′ = f(y, y ′ ). The independent variable t is missing.
It is simple to solve diffenrential equations in case (a), where function y is missing. Indeed,
if second order differential equation has the form y ′′ = f(t, y ′ ), then the equation for v = y ′
is the first order equation v ′ = f(t, v).
Example 2.1.8: Find the y solution of the second order nonlinear equation y ′′ = −2t (y ′ ) 2
with initial conditions y(0) = 2, y ′ (0) = 1.
Solution: Introduce v = y ′ . Then v ′ = y ′′ , and
v ′ = −2t v 2 ⇒ v′
= −2t
v2 ⇒
1
−
v = −t2 + c.
So, 1
y ′ = t2 − c, that is, y ′ = 1
t2 . The initial condition implies
− c
1 = y ′ (0) = − 1
c ⇒ c = −1 ⇒ y′ = 1
t2 − 1 .
⊳

58 G. NAGY – ODE july 2, 2013
Then, y =
dt
t2 + c. We integrate using the method of partial fractions,
− 1
1
t 2 − 1 =
1
(t − 1)(t + 1) =
a
(t − 1) +
b
(t + 1) .
Hence, 1 = a(t + 1) + b(t − 1). Evaluating at t = 1 and t = −1 we get a = 1
, b = −1 . So
2 2
1
t2 1
1
=
− 1 2 (t − 1) −
1
.
(t + 1)
Therefore, the integral is simple to do,
y = 1
ln |t − 1| − ln |t + 1| + c.
2
1
2 = y(0) = (0 − 0) + c.
2
We conclude y = 1
ln |t − 1| − ln |t + 1| + 2.
2
⊳
We now consider the case (b), that is, a differential equation y ′′ = f(y, y ′ ), where the
independent variable t is missing. The method to find the solution in this case is more
involved.
Theorem 2.1.11. Consider a second order differential equation y ′′ = f(y, y ′ ), and introduce
the function v(t) = y ′ (t). If the function y is invertible, then the new function ˆv(y) = v(t(y))
satisfies the first order differential equation
dˆv 1
= f(y, ˆv(y)).
dy ˆv
Proof: Notice that v ′ (t) = f(y, v(t)). Now, by chain rule
dˆv
=
dy y
dv
dt
=
dt t(y) dy t(y)
v′
y ′
=
t(y)
v′
v
Therefore, dˆv
dy
t(y)
= f y, v)
v
t(y)
1
= f(y, ˆv(y)).
ˆv
Example 2.1.9: Find a solution y to the second order equation y ′′ = 2y y ′ .
Solution: The variable t does not appear in the equation. Hence, v(t) = y ′ (t). The
equation is v ′ (t) = 2y(t) v(t). Now introduce ˆv(y) = v(t(y)). Then
dˆv
dy =
dv dt
t(y)
=
dt dy
v′
y ′
=
t(y)
v′
.
v t(y)
Using the differential equation,
dˆv 2yv
=
dy v
t(y)
⇒ dˆv
dy = 2y ⇒ ˆv(y) = y2 + c.
Since v(t) = ˆv(y(t)), we get v(t) = y 2 (t) + c. This is a separable equation,
y ′ (t)
y2 = 1.
(t) + c
Since we only need to find a solution of the equation, and the integral depends on whether
c > 0, c = 0, c < 0, we choose (for no special reason) only one case, c = 1.
dy
= dt + c0 ⇒ arctan(y) = t + c0y(t) = tan(t + c0).
1 + y2 .

G. NAGY – ODE July 2, 2013 59
Again, for no reason, we choose c 0 = 0, and we conclude that one possible solution to our
problem is y(t) = tan(t). ⊳

60 G. NAGY – ODE july 2, 2013
2.1.6. Exercises.
2.1.1.- Compute the Wronskian of the following
functions:
(a) f(t) = sin(t), g(t) = cos(t).
(b) f(x) = x, g(x) = x e x .
(c) f(θ) = cos 2 (θ), g(θ) = 1 + cos(2θ).
2.1.2.- Find the longest interval where the
solution y of the initial value problems
below is defined. (Do not try to solve
the differential equations.)
(a) t 2 y ′′ + 6y = 2t, y(1) = 2, y ′ (1) = 3.
(b) (t − 6)y ′ + 3ty ′ − y = 1, y(3) =
−1, y ′ (3) = 2.
2.1.3.- (a) Verify that y1(t) = t 2 and
y2(t) = 1/t are solutions to the differential
equation
t 2 y ′′ − 2y = 0, t > 0.
(b) Show that y(t) = a t 2 + b
is so-
t
lution of the same equation for all
constants a, b ∈ R.
2.1.4.- Can the function y(t) = sin(t 2 ) be
solution on an open interval containing
t = 0 of a differential equation
y ′′ + a(t) y ′ + b(t)y = 0,
with continuous coefficients a and b?
Explain your answer.
2.1.5.- Verify whether the functions y1, y2
below are a fundamental set for the differential
equations given below:
(a) y ′′ +4y = 0, y1(t) = cos(2t), y2(t) =
sin(2t).
(b) y ′′ − 2y ′ + y = 0, y1(t) = e t , y2(t) =
t e t .
(c) x 2 y ′′ − 2x(x + 2) y ′ + (x + 2) y =
0, y1(x) = x, y2(t) = x e x .
2.1.6.- If the Wronskian of any two solutions
of the differential equation
y ′′ + p(t) y ′ + q(t) y = 0
is constant, what does this imply about
the coefficients p and q?
2.1.7.- Find the solution y to the second order,
nonlinear equation
t 2 y ′′ + 6t y ′ = 1, t > 0.

G. NAGY – ODE July 2, 2013 61
2.2. Constant coefficients
2.2.1. Solution formulas for constant coefficients. In the previous Section we described
general properties of solutions to second order, linear, homogeneous, equations with variable
coefficients. Among them we mentioned the superposition property, how to obtain fundamental
solutions by solving appropriate initial value problems, and the evolution equation
for the Wronskian of two solutions. In this Section we focus on the particular case of constant
coefficients equations. We find every solution to differential equations in this reduced
class. We start with a useful definition before we present the main result of this Section.
Definition 2.2.1. Consider the homogeneous, constant coefficients differential equation
y ′′ + a 1y ′ + a 0 = 0.
The characteristic polynomial of the equation is
p(r) = r 2 + a1r + a0.
The characteristic equation is the equation p(r) = 0.
The following result says that the roots of the characteristic polynomial are crucial to
express the solutions of the homogeneous, constant coefficients, differential equation above.
The characteristic polynomial is a real coefficients, second degree polynomial and the general
expression for its roots is
r± = 1
.
2
−a1 ± a 2 1 − 4a0
If the discriminant (a 2 1 − 4a0) is positive, zero, or negative, then the roots of p are different
real numbers, only one real number, or a complex-conjugate pair of complex numbers. For
each case the solution of the differential equation can be expressed in different forms.
Theorem 2.2.2 (Constant coefficients). Given real constants a1, a0 consider the homogeneous,
linear differential equation on the unknown y : R → R given by
y ′′ + a1y ′ + y0 = 0. (2.2.1)
Let r± be the roots of the characteristic polynomial and let c0, c1 be arbitrary constants.
(a) If r+ = r−, real or complex, then the general solution of Eq. (2.2.1) is given by
y(t) = c0 e r-t + c1 e r-t .
(b) If r+ = r− = ˆr ∈ R, then the general solution of Eq. (2.2.1) is given by
y(t) = c 0 e ˆrt + c 1 te ˆrt .
Furthermore, given real constants t 0, y 0 and y 1, there is a unique solution to the initial value
problem given by Eq. (2.2.1) and the initial conditions y(t 0) = y 0 and y ′ (t 0) = y 1.
In the case that the characteristic polynomial has different roots, these roots can be either
real-valued of complex-valued. We present examples of solutions with real-valued roots
in this Section. In the case that the roots are complex-valued, one root is the complex
conjugate of the other, say r± = α ± βi. This is so because the characteristic polynomial
has real coefficients. In this case fundamental solutions of the differential equation are still
given by the expression of the Theorem, that is,
y1(t) = e (α+iβ)t , y2(t) = e (α−iβ)t .
We will see in the next Section that simple linear combinations of these solutions define
real-valued fundamental solutions. More often than not such solutions are useful when one
solves problems in physics or engineering.

62 G. NAGY – ODE july 2, 2013
Proof of Theorem 2.2.2: The proof has two main parts: First, we transform the original
equation into an equation simpler to solve for a new unknown; second, we solve this simpler
problem.
In order to transform the problem into a simpler one, we express the solution y as a
product of two functions, that is, y(t) = u(t)v(t). Choosing v in an appropriate way the
equation for u will be simpler to solve than the equation for y. Hence,
y = uv ⇒ y ′ = u ′ v + v ′ u ⇒ y ′′ = u ′′ v + 2u ′ v ′ + v ′′ u.
Therefore, Eq. (2.2.1) implies that
that is,
(u ′′ v + 2u ′ v ′ + v ′′ u) + a 1 (u ′ v + v ′ u) + a 0 uv = 0,
u ′′
+ a1 + 2 v′
v
We now choose the function v such that
u ′
+ a0 u v + (v ′′ + a1 v ′ ) u = 0. (2.2.2)
a1 + 2 v′
= 0
v
⇔
v′
= −a1 .
v 2
(2.2.3)
We choose a simple solution of this equation, given by
v(t) = e −a1t/2 .
Having this expression for v one can compute v ′ and v ′′ , and it is simple to check that
v ′′ + a 1v ′ = − a2 1
v. (2.2.4)
4
Introducing the first equation in (2.2.3) and Eq. (2.2.4) into Eq. (2.2.2), and recalling that
v is non-zero, we obtain the simplified equation for the function u, given by
u ′′ − k u = 0, k = a2 1
4 − a 0. (2.2.5)
Eq. (2.2.5) for u is simpler than the original equation (2.2.1) for y since in the former there
is no term with the first derivative of the unknown function.
In order to solve Eq. (2.2.5) we repeat the idea followed to obtain this equation, that
is, express function u as a product of two functions, and solve a simple problem of one of
the functions. We first consider the harder case, which is when k = 0. In this case, let us
express u(t) = e √ kt w(t). Hence,
u ′ = √ √ √
kt kt ′
ke w + e w ⇒
√ √ √ √
′′ kt kt ′ kt ′′
u = ke w + 2 ke w + e w .
Therefore, Eq. (2.2.5) for function u implies the following equation for function w
0 = u ′′ − ku = e
√ kt (2 √ k w ′ + w ′′ ) ⇒ w ′′ + 2 √ kw ′ = 0.
Only derivatives of w appear in the latter equation, so denoting x(t) = w ′ (t) we have to
solve a simple equation
Integrating we obtain w as follows,
renaming c1 = −x0/(2 √ k), we obtain
x ′ = −2 √ k x ⇒ x(t) = x0e −2√ kt , x0 ∈ R.
w ′ = x0e −2√ kt ⇒ w(t) = − x0
2 √ k e−2√ kt + c0.
w(t) = c1e −2√ √
kt kt
+ c0 ⇒ u(t) = c0e + c1e −√kt .

G. NAGY – ODE July 2, 2013 63
We then obtain the expression for the solution y = uv, given by
y(t) = c0e (− a 1 2 + √ k)t + c1e (− a 1 2 − √ k)t .
Since k = (a2 1/4 − a0), the numbers
r± = − a1
2 ± √ k ⇔ r± = 1
−a1 ±
2
a2
1 − 4a0 are the roots of the characteristic polynomial
r 2 + a1 r + a0 = 0,
we can express all solutions of the Eq. (2.2.1) as follows
y(t) = c 0e r-t + c 1e r-t , k = 0.
Finally, consider the case k = 0. Then, Eq. (2.2.5) is simply given by
u ′′ = 0 ⇒ u(t) = (c 0 + c 1t) c 0, c 1 ∈ R.
Then, the solution y to Eq. (2.2.1) in this case is given by
y(t) = (c0 + c1t) e −a1t/2 .
Since k = 0, the characteristic equation r 2 +a1 r+a0 = 0 has only one root r+ = r− = −a1/2,
so the solution y above can be expressed as
y(t) = (c0 + c1t) e r-t , k = 0.
The furthermore part follows from solving a 2 × 2 linear system for the unknowns c0 and c1.
The initial conditions for the case r- = r- are the following,
y 0 = c 0 e r-t0 + c1 e r-t0 , y1 = r+c 0 e r-t0 + r−c 1 e r-t0 .
It is not difficult to verify that this system is always solvable and the solutions are
c0 = − (r−y0 − y1) (r+ − r−) er-t0 , c1 = (r+y0 − y1) .
(r+ − r−) er-t0 The initial conditions for the case r+ = r− are the following,
y0 = (c0 + c1t0) e r-t0 , y1 = c1 e r-t0 + r+(c0 + c1t0) e r-t0 .
It is also not difficult to verify that this system is always solvable and the solutions are
c0 = y0 + t0(r+y0 − y1)
er-t0 , c1 = − (r+y0 − y0) er-t0 .
This establishes the Theorem.
2.2.2. Different real-valued roots. We now present few examples in the case that the
characteristic polynomial has two different real-valued roots.
Example 2.2.1: Find solutions to the equation
y ′′ + 5y ′ + 6y = 0. (2.2.6)
Solution: We see in Theorem 2.2.2 that at least one solution of the differential equation
above has the form y(t) = e rt . Introduce this function and its derivatives, y ′ (t) = re rt and
y ′′ (t) = r 2 e rt in the equation. The result is the characteristic equation for r, that is,
(r 2 + 5r + 6)e rt = 0 ⇔ r 2 + 5r + 6 = 0. (2.2.7)
The solutions of the characteristic equation are
r± = 1 √ 1
−5 ± 25 − 24 = (−5 ± 1) ⇒
2
2
r+ = −2,
r− = −3.

64 G. NAGY – ODE july 2, 2013
Therefore, a set of fundamental solutions to the differential equation in (2.2.6) is given by
y 1(t) = e −2t , y 2(t) = e −3t .
The superposition property in Proposition 2.1.2 implies that we have found infinitely many
solutions, called the general solution, given by
y(t) = c 1e −2t + c 2e −3t , c 1, c 2 ∈ R. (2.2.8)
Example 2.2.2: Find the solution y of the initial value problem
y ′′ + 5y ′ + 6 = 0, y(0) = 1, y ′ (0) = −1.
Solution: We know that a solution of the differential equation above is
y(t) = c 1e −2t + c 2e −3t .
We now find the constants c1 and c2 that satisfy the initial conditions above:
1 = y(0) = c 1 + c 2
−1 = y ′ (0) = −2c1 − 3c2
Therefore, the unique solution to the initial value problem is
⇒
y(t) = 2e −2t − e −3t .
c1 = 2,
c 2 = −1.
Example 2.2.3: Find the general solution y of the differential equation
2y ′′ − 3y ′ + y = 0.
Solution: We look for every solutions of the form y(t) = ert , where r is solution of the
characteristic equation
2r 2 − 3r + 1 = 0 ⇒ r = 1
⎧
√ ⎨ r1 = 1,
3 ± 9 − 8 ⇒
4
⎩ r2 = 1
2 .
Therefore, the general solution of the equation above is
y(t) = c1e t + c2e t/2 .
⊳
⊳
⊳

2.2.3. Exercises.
2.2.1.- . 2.2.2.- .
G. NAGY – ODE July 2, 2013 65

66 G. NAGY – ODE july 2, 2013
2.3. Complex roots
We have seen in the previous Section that Theorem 2.2.2 provides fundamental solutions
to the constant coefficient homogeneous equation
y ′′ + a1 y ′ + a0 y = 0, a1, a0 ∈ R. (2.3.1)
In the case (a1 − 4a0) < 0 the roots of the characteristic polynomial are different and
complex-valued, say r± = α±iβ, with α = −a1/2 and β = (a1 − a 2 1/4). The fundamental
solutions shown in Theorem 2.2.2 for this case are the complex-valued functions
˜y 1 = e (α+iβ)t , ˜y 2 = e (α−iβ)t ,
hence, the general solutions of the differential equation is
y1(t) = ˜c1e (α+iβ)t + ˜c2 e (α−iβ)t , ˜c1, ˜c2 ∈ C.
This expression for the general solution has the unfortunate property that it is not straightforward
to separate real-valued from complex-valued solutions. Such separation could be
important in physical applications. If a physical system is described by a differential equation
with real coefficients, then more often than not one is interested in finding real-valued
solutions to that equation. In the first part of this Section we provide a new set of fundamental
solutions that are real-valued. These solutions makes it trivial to separate real-valued
from complex-valued solutions. In the second part of this Section we describe the electric
current in a circuit formed by a resistance, an inductance, and a capacitor, called the RLC
circuit. This system is described by a second order, linear, constant coefficients, homogeneous
differential equation with a characteristic polynomial having complex-valued roots.
The RLC circuit is a simple example of a physical situation involving oscillations and dissipation.
Such systems are usually described by differential equations with complex-valued
roots of the characteristic polynomial.
2.3.1. Real-valued fundamental solutions. We mentioned that Theorem 2.2.2 provides
complex-valued fundamental solutions to the equation in (2.3.1) in the case that the associated
characteristic polynomial has complex roots. The following result provides real-valued
fundamental solutions in this same case.
Corollary 2.3.1 (Real-valued fundamental solutions). Assume that a1, a0 ∈ R satisfy
that a 2 1 − 4a 0 < 0, so that Theorem 2.2.2 implies that the differential equation
y ′′ + a 1 y ′ + a 0 y = 0 (2.3.2)
has the characteristic polynomial p(r) = r2 +a1r +a0 with complex-valued roots r± = α±iβ,
α = − a1 1
, β = 4a0 − a
2 2
2 1.
and the differential equation in (2.3.2) has complex-valued fundamental solutions
˜y1(t) = e (α+iβt) , ˜y2(t) = e (α−iβt) .
Then, real-valued fundamental solutions to the differential equation in (2.3.2) are given by
y1(t) = e αt cos(βt), y2(t) = e αt sin(βt).
We know from Theorem 2.2.2 that ˜y 1, ˜y 2 are fundamental set of solutions to Eq. (2.3.2).
These fundamental solutions are complex-valued. The corollary above says that there exists
real-valued fundamental solutions given by y1 and y2.
Proof of Corollary 2.3.1: Since the functions
˜y 1(t) = e (α+iβ)t , ˜y 2(t) = e (α−iβ)t ,

G. NAGY – ODE July 2, 2013 67
are solutions to the differential equation in the Corollary, then so are the functions
y1(t) = 1
˜y1(t) + ˜y2(t)
2
, y2(t) = 1
˜y1(t) − ˜y2(t)
2i
.
It is simple to check that
y1(t) = 1
2 eαte iβt + e −iβt
y2(t) = 1
2i eαte iβt − e −iβt ,
where e (α±iβ)t = eαte ±iβt . Recalling the Euler equation and its complex-conjugate,
e iβt = cos(βt) + i sin(βt), e −iβt = cos(βt) − i sin(βt),
we obtain that
y1(t) = e αt cos(βt), y2(t) = e αt sin(βt).
This establishes the Corollary.
Example 2.3.1: Find the real-valued general solution of the equation
y ′′ − 2y ′ + 6y = 0.
Solution: We first find the roots of the characteristic polynomial,
r 2 − 2r + 6 = 0 ⇒ r± = 1
√
2 ± 4 − 24 ⇒ r± = 1 ± i
2
√ 5.
We know from Theorem 2.2.2 that the complex-valued general solution of the differential
equation above can be written as follows,
y(t) = ˜c1e (1+i√ 5) t + ˜c2e (1−i√ 5) t , ˜c1, ˜c2 ∈ C.
The main information given in the Corollary above is that this general solution can also be
written the equivalent way,
y(t) = c1 cos( √ 5 t) + c2 sin( √ 5 t) e t , c1, c2 ∈ R. (2.3.3)
In order to translate from the former expression to the latter one, we only need to remember
two main relations: First, that the following equation holds
e (1±i√ 5) t = e t e ±i √ 5 t .
Second, that adding and subtracting Euler’s formula and its complex-conjugate formula
gives the relations
e i√ 5 t = cos( √ 5 t) + i sin( √ 5 t),
e −i√ 5 t = cos( √ 5 t) − i sin( √ 5 t),
i
e √ 5 t −i
+ e √ 5 t , sin( √ 5 t) = 1 i
e √ 5 t −i
− e √ 5 t ,
cos( √ 5 t) = 1
2
2i
Therefore, it is simple to show that the following relations hold,
e t cos( √ 5 t) = 1
(1+i
e
2
√ 5)t (1−i
+ e √ 5)t , e t sin( √ 5 t) = 1
2i
and so, the functions
y1(t) = e t cos( √ 5 t), y2(t) = e t sin( √ 5 t),
e (1+i √ 5)t − e (1−i √ 5)t .
are a fundamental set of solutions to the differential equation above. The real-valued general
solution is given by
y(t) = c1 cos( √ 5 t) + c2 sin( √ 5 t) e t , c1, c2 ∈ R.
⊳

68 G. NAGY – ODE july 2, 2013
Example 2.3.2: Find real-valued fundamental solutions to the equation
y ′′ + 2 y ′ + 6 y = 0.
Solution: The roots of the characteristic polynomial p(r) = r2 + 2r + 6 are
r± = 1
√ 1
√
−2 ± 4 − 24 = −2 ± −20
2
2
⇒ r± = −1 ± i √ These are complex-valued roots, with
5.
Real-valued fundamental solutions are
y
1
α = −1, β = √ 5.
y 1(t) = e −t cos( √ 5 t), y 2(t) = e −t sin( √ 5 t).
− t
e
Figure 9. The graph of the solution
y1 found in Example 2.3.1.
Example 2.3.3: Find the real-valued general solution of y ′′ + 5 y = 0.
t
Second order differential equations with
characteristic polynomials having complex
roots, like the one in this example,
describe physical processes related
to damped oscillations. An example
from physics is a pendulums with friction.
⊳
Solution: The characteristic polynomial is p(r) = r 2 + 5. Its roots are r± = ± √ 5 i. This
is the case α = 0, and β = √ 5. Real-valued fundamental solutions are
The real-valued general solution is
y1(t) = cos( √ 5 t), y2(t) = sin( √ 5 t).
y(t) = c1 cos( √ 5 t) + c2 sin( √ 5 t), c1, c2 ∈ R.
Remark: Physical processes that oscillate in time without dissipation could be described
by differential equations like the one in this example. ⊳
2.3.2. The RLC electric circuit. We describe the electric current flowing through an
electric circuit consisting of a resistance, a non-zero capacitor, and non-zero inductance as
shown in Fig. 10. A capacitor and an inductance alone can be used to produce and keep
flowing an electric current I through the circuit, from one plate of the capacitor to the other
and vice-versa, endlessly. The presence of a resistance transforms the current energy into
heat, damping the current oscillation. This system is described by an integro-differential
equation found by Kirchhoff, now called Kirchhoff’s voltage law, relating the resistance R,
capacitance C, inductance L, and the current I in a circuit as follows,
L I ′ (t) + R I(t) + 1
C
t
t0
I(s) ds = 0.

Derivate both sides above,
and divide by L,
G. NAGY – ODE July 2, 2013 69
R C L
I (t) : electric current.
Figure 10. The RLC electric circuit.
L I ′′ (t) + R I ′ (t) + 1
I(t) = 0,
C
I ′′
R
(t) + 2 I
2L
′ (t) + 1
I(t) = 0.
LC
If we introduce α = R
1
and ω = √ , then Kirchhoff’s law can be expressed as the second
2L LC
order, homogeneous, constant coefficients, differential equation
I ′′ + 2α I ′ + ω 2 I = 0.
This equation is what we need to solve our final example.
Example 2.3.4: Find real-valued fundamental solutions to I ′′ + 2α I ′ + ω 2 I = 0, where
α = R/(2L), ω 2 = 1/(LC), in the cases (a), (b) below.
Solution: The roots of the characteristic polynomial is p(r) = r2 + 2αr + ω2 , are given by
r± = 1
−2α ± 4α2 − 4ω2 2
⇒ r± = −α ± α2 − ω2 .
Case (a): R = 0. This implies α = 0, so r± = ±iω. Therefore,
I1(t) = cos(ωt), I2(t) = sin(ωt).
Remark: When the circuit has no resistance, the current oscillates without dissipation.
Case (b): R < 4L/C. This implies
R 2 < 4L
C
R2 1
⇔ <
4L2 LC ⇔ α2 < ω 2 .
Therefore, the characteristic polynomial has complex roots r± = −α ± i √ ω 2 − α 2 , hence
the fundamental solutions are
I1(t) = e −αt cos ω 2 − α 2 t ,
I 2(t) = e −αt sin ω 2 − α 2 t .
Therefore, the resistance R damps the current
oscillations produced by the capacitor
and the inductance. ⊳
I
1
− t
e
Figure 11. The electric current
I as function of time.
t

70 G. NAGY – ODE july 2, 2013
2.3.3. Exercises.
2.3.1.- . 2.3.2.- .

G. NAGY – ODE July 2, 2013 71
2.4. Repeated roots
In Section 2.2 we have proven Theorem 2.2.2, which states every solution to the equation
y ′′ + a 1 y ′ + a 0 y = 0
for every possible value of the coefficients a1, a0 ∈ R. In particular, in the case that a 2 1 = 4a0,
that is, the characteristic polynomial has the repeated root r+ = r− = −a 1/2, the Theorem
states that a fundamental set of solutions is given by
y 1(t) = e r-t , y 2(t) = te r-t . (2.4.1)
In the first part of this Section we present two new proofs for the repeated roots part of
Theorem 2.2.2. The first proof contains a nice argument showing that the fundamental
solutions for the case of repeated roots can be obtained from a limit procedure on solutions
to equations with complex roots. More precisely, we show that solutions of equations
with complex roots having small imaginary part are close to solutions of equations with
appropriate repeated roots. We then present a third proof of the repeated root part in
Theorem 2.2.2. Although this is essentially the original proof given in Section 2.2, here we
emphasize its main idea. The reason for highlighting this idea becomes clear in the second
part of this Section. There we generalize this idea, called the reduction of order method, to
differential equations with variable coefficients.
2.4.1. Fundamental solutions for repeated roots. We start with a simple example
showing how we use the repeated root part of Theorem 2.2.2.
Example 2.4.1: Find the solution to the initial value problem
9y ′′ + 6y ′ + y = 0, y(0) = 1, y ′ (0) = 5
3 .
Solution: The characteristic polynomial is p(r) = 9r2 + 6r + 1, with root
r± = 1 √
−6 ± 36 − 36 ⇒ r+ = r− = −
18
1
3 .
Therefore, Theorem 2.2.2 says that the general solution has the form
y(t) = c1e −t/3 + c2te −t/3 .
Since its derivative is given by
y ′ (t) = − c1
3 e−t/3
+ c2 1 − t
e
3
−t/3 ,
then, the initial conditions imply that
1 = y(0) = c1, 5
3 = y′ (0) = − c1 3 + c ⎫
⎬
⎭ 2
⇒ c1 = 1, c2 = 2.
So, the solution to the initial value problem above is: y(t) = (1 + 2t) e −t/3 . ⊳
Second proof of repeated root part of Theorem 2.2.2: We now show that the solution
y 2 in Eq. (2.4.1) is close to solutions of an appropriate differential equation having complex
roots in the case that the imaginary part is small enough. Indeed, consider the differential
equation
˜y ′′ + ã 1 ˜y ′ + ã 0 ˜y = 0,
where ã1 − 4ã0 < 0. We know that real-valued fundamental solutions are given by
˜y1 = e α t cos(βt), ˜y2 = e α t sin(βt),

72 G. NAGY – ODE july 2, 2013
where α = −ã 1/2 and β = ã 0 − ã 2 1/4, as usual. We now study what happen to these
solutions ˜y1 and ˜y2 in the following limit: The variable t is held constant, α is held constant,
and β → 0. The last two conditions are conditions on the equation coefficients, ã1, ã0.
Since cos(βt) → 1 as β → 0 with t fixed, then keeping α fixed too, we obtain
Since sin(βt)
βt
˜y1β(t) = e α t cos(βt) −→ e α t = y1(t).
→ 1 as β → 0 with t constant, that is, sin(βt) → βt, we conclude that
˜y2(t)
β
= sin(βt)
β
e α t = sin(βt)
βt
t e α t −→ t e α t = y2(t).
The calculation above says that the function ˜y2(t) α t
/β is close to the function y2(t) = t e
in the limit mentioned above. This calculation only suggests that y2(t) = t y1(t) could be
solutions of the equation having repeated roots in its characteristic polynomial. It gives a
hint on what could be the solution. One has to introducing this function y2 in the differential
equation to prove that this function y2 is indeed a solution. Since y2 is not proportional to
y1, one then concludes the functions y1, y2 are a fundamental set for the differential equation
in with repeated roots. This establishes part (b) in Theorem 2.2.2.
We now repeat one more time the last part of the proof of Theorem 2.2.2 in order to
highlight its main idea. It turns out that this idea can be generalized from equations with
constant coefficients to equations with variable coefficients, and it is called the reduction of
order method. This generalization occupies the last part of this Section.
Third proof of repeated root part of Theorem 2.2.2: Since a2 1 = 4a0, there is only
one root to the characteristic polynomial p(r) = r2 + a1r + a0 given by r1 = −a1/2. This
implies that one solution of the differential equation y ′′ + a1 y ′ + a0 y = 0 is given by
y 1(t) = e r1t .
The main idea in the reduction of order method is to look for a second solution to the
differential equation as a product of a function times the known solution, that is,
y2(t) = v(t) y1(t) ⇒ y2(t) = v(t) e r1t .
Introducing this function y2 into the differential equation one obtains a differential equation
for the new function v. First compute
y ′ 2 = (v ′ + r1 v) e r1t , y ′′
2 = (v ′′ + 2r1 v ′ + r 2 1 v) e r1t .
As we said, introduce y2 into the differential equation and get the equation for v,
0 = (v ′′ + 2r1 v ′ + r 2 1 v) + a1(v ′ + r1 v) + a0 v,
= v ′′ + (2r 1 + a 1) v ′ + (r 2 1 + a 1r + a 0) v.
However, the equation for v above is simpler than the original equation for y2, since y1 = e r1t
is a solution of the original differential equation, which in this case means both that
r 2 1 + a1r1 + a0 = 0 and 2r1 + a1 = 0.
So, the equation for the function v is simple and given by
v ′′ = 0.
The solutions are given in terms of two arbitrary constants c 1 and c 2 as follows
v(t) = c 1 + c 2t ⇒ y 2(t) = c 1e r1t + c2te r1t .

G. NAGY – ODE July 2, 2013 73
Since the constants c 1 and c 2 are arbitrary, choosing c 1 = 0 and c 2 = 1 one obtains the
following set of solutions to the original differential equation:
y1(t) = e r1t , y2(t) = te r1t .
This establishes part (b) in Theorem 2.2.2.
2.4.2. Reduction of order method. This same idea used in the proof above, the reduction
of order method, can be generalized to variable coefficient equations.
Theorem 2.4.1 (Reduction of order). Given continuous functions p, q : (t 1, t 2) → R,
let y1 : (t1, t2) → R be a solution of the differential equation
If a function v : (t 1, t 2) → R is solution of
y ′′ + p(t) y ′ + q(t) y = 0. (2.4.2)
y 1(t) v ′′ + 2y ′ (t) + p(t)y 1(t) v ′ = 0, (2.4.3)
then the functions y 1 and y 2 = v y 1 are fundamental solutions to Eq. (2.4.2).
The reason for the name reduction of order method is that the equation for the function
v involves v ′′ and v ′ but does not involve v itself. So this equation is a first order equation
for the unknown w = v ′ .
Proof of Theorem 2.4.1: If we express y2 = vy1, then the differential equation for y
implies an equation for v, as the following calculation shows. First, compute y ′ 2 and y ′′
2 ,
y ′ 2 = v ′ y1 + v y ′ 1, y ′′
2 = v ′′ y1 + 2v ′ y ′ 1 + v y ′′
1 .
Introduce this information introduced the differential equation,
0 = (v ′′ y1 + 2v ′ y ′ 1 + v y ′′
1 ) + p (v ′ y1 + v y ′ 1) + qv y1
= y1 v ′′ + (2y ′ 1 + p y1) v ′ + (y ′′
1 + p y ′ 1 + q y1) v.
The function y1 is solution to the differential original differential equation,
y ′′
1 + p y ′ 1 + q y1 = 0,
then, the equation for v is given by Eq. (2.4.3), that is,
y1 v ′′ + (2y ′ 1 + p y1) v ′ = 0.
We now need to show that y1 and y2 = vy1 are linearly independent.
Wy1y2 =
y1 vy1
y ′ 1 (v ′ y1 + vy ′
1)
We need to find v ′ . Denote w = v ′ , so
= y1(v ′ y1 + vy ′ 1) − vy1y ′ 1 ⇒ Wy1y2 = v′ y 2 1.
y1 w ′ + (2y ′ 1 + p y1) w = 0 ⇒ w′
w = −2y′ 1
Let P be a primitive of p, that is, P ′ (t) = p(t), then
ln(w) = −2 ln(y1) − P ⇒ w = e [ln(y−2
y1
− p.
1 )−P ] ⇒ w = y −2
1
e −P .
We obtain v ′ y 2 1 = e −P , hence Wy1y2 = e−P , which is non-zero. We conclude that y1 and
y 2 = vy 1 are linearly independent. This establishes the Theorem.

74 G. NAGY – ODE july 2, 2013
Example 2.4.2: Find a second solution y 2 linearly independent to the solution y 1(t) = t of
the differential equation
t 2 y ′′ + 2ty ′ − 2y = 0.
Solution: We look for a solution of the form y 2(t) = t v(t). This implies that
So, the equation for v is given by
y ′ 2 = t v ′ + v, y ′′
2 = t v ′′ + 2v ′ .
0 = t 2 t v ′′ + 2v ′ + 2t t v ′ + v − 2t v
= t 3 v ′′ + (2t 2 + 2t 2 ) v ′ + (2t − 2t) v
= t 3 v ′′ + (4t 2 ) v ′ ⇒ v ′′ + 4
t v′ = 0.
Notice that this last equation is precisely Eq. (2.4.3), since in our case we have
y1 = t, p(t) = 2
t ⇒ t v′′ + 2 + 2
t t v ′ = 0.
The equation for v is a first order equation for w = v ′ , given by
w ′
= −4
w t ⇒ w(t) = c1t −4 , c1 ∈ R.
Therefore, integrating once again we obtain that
v = c2t −3 + c3, c2, c3 ∈ R,
and recalling that y 2 = t v we then conclude that
y2 = c2t −2 + c3t.
Choosing c2 = 1 and c3 = 0 we obtain that y2(t) = t−2 . Therefore, a fundamental solution
set to the original differential equation is given by
y1(t) = t, y2(t) = 1
.
t2 ⊳

2.4.3. Exercises.
2.4.1.- . 2.4.2.- .
G. NAGY – ODE July 2, 2013 75

76 G. NAGY – ODE july 2, 2013
2.5. Undetermined coefficients
The method of undetermined coefficients is a trial-and-error procedure to find solutions
of non-homogeneous differential equations. Recall that a differential equation of the form
y ′′ + p(t) y ′ + q(t) y = f(t) (2.5.1)
is called non-homogeneous iff the function f is not identically zero; see Definition 2.1.1.
The method of undetermined coefficients is then, given a particular form of the function
f, propose a clever guess for the solution y, guess that depends on some free parameters,
or yet undetermined coefficients. Introduce this guess y into the differential equation and
determine the free parameters.
2.5.1. Operator notation. Given functions p, q : (t1, t2) → R, we denote by L(y) the
left-hand side in Eq. (2.5.1), that is
L(y) = y ′′ + p(t) y ′ + q(t) y. (2.5.2)
The function L acting on the function y is called an operator. Therefore, Eq. (2.5.1) can be
written as L(y) = f, while the homogeneous equation is simply L(y) = 0.
It is also convenient to emphasize that the superposition property proven in Proposition
2.1.2 is a consequence of the fact that the operator L is a linear function of y. This is
stated in the following result.
Theorem 2.5.1 (Linearity of L). For every continuously differentiable pair of functions
y 1, y 2 : (t 1, t 2) → R and every c 1, c 2 ∈ R the operator in (2.5.2) satisfies that
L(c1y1 + c2y2) = c1L(y1) + c2L(y2). (2.5.3)
Proof of Theorem 2.5.1: This is a straightforward calculation:
L(c 1y 1 + c 2y 2) = (c 1y 1 + c 2y 2) ′′ + p(t) (c 1y 1 + c 2y 2) ′ + q(t) (c 1y 1 + c 2y 2).
Re-order terms in the following way,
L(c1y1 + c2y2) = c1y ′′
1 + p(t) c1y ′ 1 + q(t) c1y1
+ c2y ′′
2 + p(t) c2y ′
2 + q(t) c2y2 .
Introduce the definition of L back on the right-hand side. We then conclude that
L(c1y1 + c2y2) = c1L(y1) + c2L(y2).
This establishes the Theorem.
It is now simple to see that the superposition property follows from the linearity of the
operator L. Indeed, for every functions y 1, y 2, solutions of the homogeneous equations
L(y 1) = 0 and L(y 2) = 0 holds that their linear combination (c 1y 1 + c 2y 2) is also solution of
the homogeneous equation, since
L(c 1y 1 + c 2y 2) = c 1L(y 1) + c 2L(y 2) = 0.
Finally, we state two simple results we need later on. Their proof is a straightforward
consequence of Eq. (2.5.3), and they are left as exercise for the reader.
Theorem 2.5.2 (General solution). Given functions p, q, f, let L(y) = y ′′ + p y ′ + q y.
If the functions y 1 and y 2 are fundamental solutions of the homogeneous equations
L(y 1) = 0, L(y 2) = 0,
and yp is any solution of the non-homogeneous equation
L(yp) = f, (2.5.4)
then any other solution y of the non-homogeneous equation above is given by
y(t) = c1y1(t) + c2y2(t) + yp(t), c1, c2 ∈ R. (2.5.5)

G. NAGY – ODE July 2, 2013 77
We just mention that the expression for y in Eq. (2.5.5) is called the general solution of
the non-homogeneous Eq. (2.5.4). Finally, we present a second result that uses the linearity
property of the operator L. This result is useful to simplify solving certain differential
equations.
Theorem 2.5.3 (Source simplification). Given functions p, q, let L(y) = y ′′ + p y ′ + q y.
If the function f can be written as f(t) = f1(t) + · · · + fn(t), with n 1, and if there exists
functions yp1 , · · · , ypn such that
L(ypi ) = fi, i = 1, · · · , n,
then the function yp = yp1 + · · · + ypn satisfies the non-homogeneous equation
L(yp) = f.
2.5.2. The undetermined coefficients method. The method of undetermined coefficients
is used to solve the following type of problems: Given a constant coefficients linear
operator L(y) = y ′′ +a1y ′ +a0y, with a1, a2 ∈ R, find every solution of the non-homogeneous
differential equation
L(y) = f.
A summary of the undetermined coefficients method is the following:
(1) Find the general solution to the homogeneous equation L(yh) = 0.
(2) If the source function f has the form f = f1+· · ·+fn, with n 1, then look for solutions
yp1 , · · · , ypn , to the equations L(ypi ) = fi, with i = 1, · · · , n. Once the functions ypi are
found, then construct yp = yp1 + · · · + ypn .
(3) Given the source functions fi, guess the solutions ypi following the Table 1 below.
(4) If any guessed function ypi satisfies the homogeneous equation L(ypi) = 0, then change
the guess to the function tsypi , with s 1 sufficiently large such that L(tsypi ) = 0.
(5) Impose the equation L(ypi) = fi to find the undetermined constants k1, · · · , kn given in
the table below. Then, compute yp = yp1 + · · · + ypn .
(6) The solution to the original problem is then given by y(t) = yh(t) + yp(t).
In the rest of the Section we use this method in several examples.
Example 2.5.1: Find all solutions to the inhomogeneous equation
y ′′ − 3y ′ − 4y = 3e 2t .
Solution: Form the problem we get that L(y) = y ′′ − 3y ′ − 4y and f(t) = 3e 2t .
(1): Find all solutions yh to the homogeneous equation L(yh) = 0. The characteristic
equation is
r 2 − 3r − 4 = 0 ⇒ r 1 = 4, r 2 = −1, ⇒ yh(t) = c 1 e 4t + c 2 e −t .
(2): Trivial in our case. The source function f(t) = 3e 2t cannot be simplified into a sum of
simpler functions.
(3): Table says: For f(t) = 3e 2t guess yp(t) = k e 2t . The constant k is the undetermined
coefficient we must find.
(4): Trivial here, since L(yp) = 0. Therefore, we do not modify our guess. (Recall: L(yh) = 0
iff yh(t) = c 1 e 4t + c 2 e −t .)

78 G. NAGY – ODE july 2, 2013
(5): Introduce yp into L(yp) = f and find k.
(2 2 − 6 − 4)ke 2t = 3e 2t ⇒ −6k = 3 ⇒ k = − 1
2 .
Notice that the guessed solution must be proportional to the exponential e 2t in order to
cancel out the exponentials in the equation above. We have obtained that
yp(t) = − 1
2 e2t .
(6): The general solution to the inhomogeneous equation is then given by
y(t) = c1e 4t + c2e −t − 1
2 e2t .
fi(t) (K, m, a, b, given.) ypi (t) (Guess) (k not given.)
Ke at ke at
Kt m kmt m + km−1t m−1 + · · · + k0
K cos(bt) k1 cos(bt) + k2 sin(bt)
K sin(bt) k 1 cos(bt) + k 2 sin(bt)
Kt m e at e at (kmt m + · · · + k 0)
Ke at cos(bt) e at k 1 cos(bt) + k 2 sin(bt)
KKeat sin(bt) eatk1 cos(bt) + k2 sin(bt)
Ktm
cos(bt)
kmtm
+ · · · + k0 a1 cos(bt) + a2 sin(bt)
Ktm
sin(bt)
kmtm
+ · · · + k0 a1 cos(bt) + a2 sin(bt)
Table 1. List of sources fi and solutions ypi to the equation L(ypi ) = fi.
Example 2.5.2: Find all solutions to the inhomogeneous equation
y ′′ − 3y ′ − 4y = 3e 4t .
Solution: We have the same left hand side as in the example above, so the first step,
finding the solutions to the homogeneous equation, is the same as in the example above:
yh(t) = c1e 4t + c2e −t .
The second step is again trivial, since the source function f(t) = 3e 4y cannot be simplified
into a sum of simpler functions. The third step says for the source f(t) = 3 e 4t guess
yp(t) = k e 4t . However, step four tells us that we have to change our guess, since yp = k yh1
is a solution of the homogeneous equation. Step four says we have to change our guess to
yp(t) = kt e 4t .
Step five is to introduce the guess into L(yp) = f. We need to compute
y ′ p = (1 + 4t) ke 4t , y ′′
p = (8 + 16t) ke 4t ⇒ 8 − 3 + (16 − 12 − 4)t ke 4t = 3e 4t ,
⊳

therefore, we get that
G. NAGY – ODE July 2, 2013 79
5k = 3 ⇒ k = 3
5 ⇒ yp(t) = 3
5 te4t .
The general solution of the non-homogeneous equation is
y(t) = c 1e 4t + c 2e −t + 3
5 te4t .
Example 2.5.3: Find all the solutions to the inhomogeneous equation
y ′′ − 3y ′ − 4y = 2 sin(t).
Solution: We again have the same left hand side as in the example above, so the first step,
finding the solutions to the homogeneous equation, is the same as in the example above:
yh(t) = c1e 4t + c2e −t .
The second step is still not needed, the third step says that for the source f(t) = 2 sin(t) the
guess must be yp(t) = k1 sin(t) + k2 cos(t). Since our guess is not a solution to the homogeneous
equation, we proceed to step five and we introduce our guess into the differential
equation. We first compute
y ′ p = k 1 cos(t) − k 2 sin(t), y ′′
p = −k 1 sin(t) − k 2 cos(t),
and then we obtain
[−k 1 sin(t) − k 2 cos(t)] − 3[k 1 cos(t) − k 2 sin(t)] − 4[k 1 sin(t) + k 2 cos(t)] = 2 sin(t).
Re-ordering terms in the expression above we get
(−5k 1 + 3k 2) sin(t) + (−3k 1 − 5k 2) cos(t) = 2 sin(t).
The last equation must hold for all t ∈ R. In particular, it must hold for t = π/2 and for
t = 0. At these two points we obtain, respectively,
⎧
⎪⎨
−5k1 + 3k2 = 2,
k1 = −
⇒
−3k1 − 5k2 = 0, ⎪⎩
5
17 ,
k2 = 3
17 .
So the particular solution to the inhomogeneous equation is given by
yp(t) = 1
−5 sin(t) + 3 cos(t) ,
17
and the general solution is
−5 sin(t) + 3 cos(t) .
y(t) = c1e 4t + c2e −t + 1
17
Example 2.5.4: We provide few more examples of non-homogeneous equations and the
appropriate guesses for the particular solutions.
(a) For y ′′ − 3y ′ − 4y = 3e 2t sin(t), guess, yp(t) = k1 sin(t) + k2 cos(t) e 2t .
(b) For y ′′ − 3y ′ − 4y = 2t 2 e 3t , guess, yp(t) = k2t 2 + k1t + k0
(c) For y ′′ − 3y ′ − 4y = 2t 2 e 4t , guess, yp(t) = t k2t 2 + k1t + k0
e 3t .
e 4t .
(d) For y ′′ − 3y ′ − 4y = 3t sin(t), guess, yp(t) = (1 + k1t) k2 sin(t) + k3 cos(t) .
⊳
⊳

80 G. NAGY – ODE july 2, 2013
2.5.3. Exercises.
2.5.1.- . 2.5.2.- .

G. NAGY – ODE July 2, 2013 81
2.6. Variation of parameters
We describe a method to find a solution to a non-homogeneous differential equation with
variable coefficients with a continuous but otherwise arbitrary source function,
y ′′ + p(t) y ′ + q(t) y = f(t),
in the case that one happens to know a two linearly independent solutions to the associated
homogeneous equation. This way to find solutions to non-homogeneous differential equations
is called the variation of parameters method and is summarized in Theorem 2.6.1 below.
The variation of parameter method can be applied to more general equations than the
undetermined coefficients method studied in the previous Section. However, the former
usually takes more time to implement than the latter simpler method.
2.6.1. The variation of parameters method. The whole method can be summarized in
the following statement.
Theorem 2.6.1 (Variation of parameters). Let p, q, f : (t1, t2) → R be continuous
functions, let y1, y2 : (t1, t2) → R be fundamental solutions to the homogeneous equation
y ′′ + p(t) y ′ + q(t) y = 0,
and let Wy1y2 be the Wronskian of y1 and y2. If the functions u1 and u2 are defined by
u1(t) = − y2(t)f(t) Wy1y2 (t) dt, u2(t)
y1(t)f(t)
=
dt, (2.6.1)
Wy1y2 (t)
then a particular solution to the non-homogeneous equation y ′′ + p(t) y ′ + q(t) y = f(t) is
yp = u 1y 1 + u 2y 2.
The result above is more general than the trial-and-error method studied earlier since
Theorem 2.6.1 applies to any continuous source function f, not only to those functions that
appear in Table 1. As in the undetermined coefficients method, the variation of parameters
method also requires to know two solutions to the homogeneous equation in order to solve
the non-homogeneous one. One can say that the proof of Theorem 2.6.1 is based in two new
ideas. While the first idea is a good one, the second idea is much better.
Proof of Theorem 2.6.1: The problem is to find a solution of an inhomogeneous equation
y ′′ + p(t) y ′ + q(t) y = f(t) (2.6.2)
knowing that there exist fundamental solutions y1 and y2 to the homogeneous equation
y ′′ + p(t) y ′ + q(t) y = 0.
One can say that the first idea to solve this inhomogeneous equation comes from the reduction
of order method. In that case, one finds a second solution y 2 to an homogeneous
equation if one already knows another solution y 1, by proposing y 2 = uy 1. One then hopes
that the equation for u will be simpler than the original equation for y2, since y1 is solution
of the homogeneous equation. We now adapt this idea to our situation as follows: we
propose a form of the solution yp of the form
yp = u1y1 + u2y2,
and we hope that the equation for u 1 and u 2 will be simpler than the original equation,
since y 1 and y 2 are solutions to the homogeneous equation. We introduce this expression
for yp into the original differential equation. We must first compute the derivatives
y ′ p = u ′ 1y 1 + u 1y ′ 1 + u ′ 2y 2 + u 2y ′ 2, y ′′
p = u ′′
1 y 1 + 2u ′ 1y ′ 1 + u 1y ′′
1 + u ′′
2 y 2 + 2u ′ 2y ′ 2 + u 2y ′′
2 .

82 G. NAGY – ODE july 2, 2013
After some reordering of terms, one obtains that Eq. (2.6.2) has the form
f = u ′′
1 y 1 + u ′′
2 y 2 + 2(u ′ 1y ′ 1 + u 2y ′ 2) + p (u ′ 1y 1 + u ′ 2y 2)
+ u1 (y ′′
1 + p y ′ 1 + q y1) + u2 (y ′′
2 + p y ′ 2 + q y2)
The functions y1 and y2 are solutions to the homogeneous equations
y ′′
1 + p y ′ 1 + q y1 = 0, y ′′
2 + p y ′ 2 + q y2 = 0,
so we obtain that u1 and u2 must be solution of a simpler equation that the one above,
given by
f = u ′′
1 y1 + u ′′
2 y2 + 2(u ′ 1y ′ 1 + u ′ 2y ′ 2) + p (u ′ 1y1 + u ′ 2y2). (2.6.3)
The second idea is the following: Look for functions u1 and u2 that satisfy the extra equation
u ′ 1y1 + u ′ 2y2 = 0. (2.6.4)
Any
solution of this extra equation must satisfy that its derivative is also zero, that is,
′ u 1y1 + u ′ ′
2y2 = 0, and this last equation is explicitly given by
u ′′
1 y1 + u ′′
2 y2 + (u ′ 1y ′ 1 + u ′ 2y ′ 2) = 0. (2.6.5)
Using Eqs. (2.6.4) and Eq. (2.6.5) into the original Eq. (2.6.3) we obtain that
u ′ 1y ′ 1 + u ′ 2y ′ 2 = f. (2.6.6)
Therefore, instead of solving the original equation in (2.6.3) we find solutions u1 and u2 for
the system given by Eqs. (2.6.4) and (2.6.6). This system is simple to solve since it is an
algebraic system of two equations for the two unknowns u ′ 1 and u ′ 2. From Eq. (2.6.4) we
compute u ′ 2 and we introduce it into Eq. (2.6.6), as follows,
u ′ 2 = − y1
u
y2 ′ 1 ⇒ u ′ 1y ′ 1 − y1y ′ 2
u
y2 ′ 1 = f ⇒ u ′ ′ y 1y2 − y1y
1
′ 2
y2 Since Wy1y2 = y1y ′ 2 − y ′ 1y2, we obtain that
u ′ 1 = − y2f
Wy1y2
⇒ u ′ 2 = y1f
Wy1y2
.
= f.
Integrating in the variable t we obtain the expressions in Eq. (2.6.1), if we choose the
integration constants to be zero. Then yp = u 1y 1 + u 2y 2 is a solution to the inhomogeneous
differential equation in Theorem (2.6.1). Notice that the constant of integration can always
be chosen to be zero, because of the following argument: Denote by u1 and u2 the functions
in Eq. (2.6.1), and given any real numbers c1 and c2 define
ũ1 = u1 + c1, ũ2 = u2 + c2.
Then the corresponding solution ˜yp is given by
˜yp = ũ 1y 1 + ũ 2y 2 = u 1y 1 + u 2y 2 + c 1y 1 + c 2y 2 ⇒ ˜yp = yp + c 1y 1 + c 2y 2.
That is, the two solutions ˜yp and yp differ by a solution to the homogeneous differential
equation, so both are solutions to the inhomogeneous equation. One is then free to chose
the constants c 1 and c 2 in any way. We choose them to be zero. This establishes the
Theorem.
Example 2.6.1: Find the general solution of the non-homogeneous equation
y ′′ − 5y ′ + 6y = 2e t .

G. NAGY – ODE July 2, 2013 83
Since the construction of yp by Theorem 2.6.1 requires to know fundamental solutions to
the homogeneous problem, let us find these solutions first. Since the equation has constant
coefficients, we compute the characteristic equation,
r 2 − 5r + 6 = 0 ⇒ r = 1
√
5 ± 25 − 24 ⇒
2
So, the functions y1 and y2 in Theorem 2.6.1 are in our case given by
y 1(t) = e 3t , y 2(t) = e 2t .
The Wronskian of these two functions is given by
r1 = 3,
r 2 = 2.
Wy1y2 (t) = (e3t )(2e 2t ) − (3e 3t )(e 2t ) ⇒ Wy1y2 (t) = −e5t .
We are now ready to compute the functions u1 and u2. Notice that Eq. (2.6.1) the following
differential equations
u ′ 1 = − y2f
, u ′ 2 = y1f
.
So, the equation for u1 is the following,
Wy1y2
Wy1y2
u ′ 1 = −e 2t (2e t )(−e −5t ) ⇒ u ′ 1 = 2e −2t ⇒ u1 = −e −2t ,
u ′ 2 = e 3t (2e t )(−e −5t ) ⇒ u ′ 2 = −2e −t ⇒ u2 = 2e −t ,
where we have chosen the constant of integration to be zero. Therefore, the particular
solution we are looking for is given by
yp = (−e −2t )(e 3t ) + (2e −t )(e 2t ) ⇒ yp = e t .
Then, the general solution to the inhomogeneous equation above is given by
y(t) = c1e 3t + c2e 2t + e t
c1, c2 ∈ R.
Example 2.6.2: Find a particular solution to the differential equation
t 2 y ′′ − 2y = 3t 2 − 1,
knowing that y1 = t 2 and y2 = 1/t are solutions to the homogeneous equation t 2 y ′′ −2y = 0.
Solution: We first rewrite the inhomogeneous equation above in the form given in Theorem
2.6.1, that is, we must divide the whole equation by t 2 ,
y ′′ − 2 1
1
y = 3 − . ⇒ f(t) = 3 −
t2 t2 t
We now proceed to compute the Wronskian
Wy1y2 (t) = (t2
−1
)
t2
1
− (2t)
t
2 .
⇒ Wy1y2 (t) = −3.
We now use the equation in (2.6.1) to obtain the functions u1 and u2, that is,
u ′ 1 = − 1
t
3 − 1
t 2
1
−3
= 1 1
−
t 3 t−3 ⇒ u1 = ln(t) + 1
6 t−2 ,
u ′ 2 = (t 2 )
3 − 1
t2
1
−3
= −t 2 + 1
3 ⇒ u 2 = − 1
3 t3 + 1
3 t.
⊳

84 G. NAGY – ODE july 2, 2013
Therefore, a particular solution to the inhomogeneous equation above is ˜yp = u1y1 + u2y2, that is,
˜yp =
ln(t) + 1
6 t−2(t
2 ) + 1
3 (−t3 + t)(t −1 )
= t 2 ln(t) + 1 1
−
6 3 t2 + 1
3
= t 2 ln(t) + 1 1
−
2 3 t2
= t 2 ln(t) + 1 1
−
2 3 y1(t).
However, a simpler expression for a solution of the inhomogeneous equation above is
yp = t 2 ln(t) + 1
2 .
Remark: Sometimes it could be difficult to remember the formulas for functions u1 and u2
in (2.6.1). In such case one can always go back to the place in the proof of Theorem 2.6.1
where these formulas come from, that is, the system
u ′ 1y1 + u ′ 2y2 = 0
u ′ 1y ′ 1 + u ′ 2y ′ 2 = f.
The system above could be simpler to remember than the equations in (2.6.1). We end this
Section using the equations above to solve the problem in Example 2.6.2. Recall that the
solutions to the homogeneous equation in Example 2.6.2 are y1(t) = t 2 , and y2(t) = 1/t,
while the source function is f(t) = 3 − 1/t 2 . Then, we need to solve the system
t 2 u ′ 1 + u ′ 1
2
2t u ′ 1 + u ′ 2
= 0,
t
(−1)
t2 1
= 3 −
t
This is an algebraic linear system for u ′ 1 and u ′ 2. Those are simple to solve. From the equation
on top we get u ′ 2 in terms of u ′ 1, and we use that expression on the bottom equation,
u ′ 2 = −t 3 u ′ 1 ⇒ 2t u ′ 1 + t u ′ 1 = 3 − 1
t2 ⇒ u′ 1 = 1 1
− .
t 3t3 Substitue back the expression for u ′ 1 in the first equation above and we get u ′ 2. We get,
u ′ 1 = 1 1
−
t 3t3 u ′ 2 = −t 2 + 1
3 .
We should now integrate these functions to get u1 and u2 and then get the particular solution
˜yp = u1y1 + u2y2. We do not repeat these calculations, since they are done Example 2.6.2.
2 .
⊳

2.6.2. Exercises.
2.6.1.- . 2.6.2.- .
G. NAGY – ODE July 2, 2013 85

86 G. NAGY – ODE july 2, 2013
Chapter 3. Power series solutions
This Chapter is focused to find solutions to second order linear differential equations
having variable coefficients. We use power series representations to find the solution. This
solutions is usually defined only in a neighborhood of a point where the power series is
centered at. In Section 3.1 we study the simplest case, when the power series is centered
at a regular point of the equation. A point is regular iff the coefficient that multiplies
the second derivative of the unknown does not vanishes there. In Section 3.2 we study a
particular type of variable coefficients equations with singular points, called Euler differential
equations. We find solutions to Euler equations in a neighborhood of their singular points.
In Section 3.3 we combine the results in the previous two Sections to find solutions to a
large class of equations, called equations with regular-singular points, which include both
Euler differential equations and regular point equations.
3.1. Regular points
In this Section we describe an idea to find solutions of variable coefficients equations
P (x) y ′′ + Q(x) y ′ + R(x) y = 0, (3.1.1)
where the functions P , Q, and R are continuous. The main idea is to assume that the
solution admits a power series representation around a point x0 ∈ R of the form
∞
y(x) = an(x − x0) n . (3.1.2)
n=0
In the case that this power series assumption is true, we can obtain the solution by simply
introducing the expansion (3.1.2) into the differential equation in (3.1.1) and then obtaining
the coefficients an that solve this equation. The power series representation of the solution
defined in this way will be defined in an interval around x 0, where the radius of convergence
can be determined after the solution is obtained. Notice that the assumption we have done
in Eq. (3.1.2) implies that the solution y is well-defined at x0, since y(x0) = a0.
In the case that the power series assumption in Eq. (3.1.2) is not true, when we solve for
the coefficients an we obtain a contradiction. There exist equations where this power series
assumption for their solutions is in fact not true, for example the Euler differential equation
(x − 2) 2 y ′′ + 6(x − 2) y ′ + 6 y = 0. (3.1.3)
We will see in the next Section that this equation has two linearly independent solutions
1
y1(x) =
(x − 2) 2 , y 1
2(x) = .
(x − 2) 3
Both solutions above diverge at x0 = 2, so if one choose to find solutions of the equation
above around x0 = 2 using a power series expansion for the solution of the form given
in (3.1.2), one is going to arrive to a contradiction. Equations similar to (3.1.3) will be
studied in later sections.
3.1.1. Review of power series. We present basic concepts about the power series representation
of functions that we will need to find solutions to differential equations. We review
only the main results without giving their proofs, which can be found in standard Calculus
textbooks. We start with the definition of a power series representation of a function.
Definition 3.1.1. The power series of a function y : R → R centered at x0 ∈ R is
∞
y(x) = an (x − x0) n .
n=0

G. NAGY – ODE July 2, 2013 87
Example 3.1.1: We show few examples of power series representation of functions:
1
(a)
1 − x =
∞
x n = 1 + x + x 2 + · · · . Here x0 = 0 and |x| < 1.
(b) e x =
n=0
∞
n=0
n=0
x n
n!
= 1 + x + x2
2! + · · · . Here x 0 = 0 and x ∈ R.
(c) The Taylor series of y : R → R centered at x0 ∈ R is
∞ y
y(x) =
(n) (x0)
n!
(x − x 0) n ⇒ y(x) = y(x 0) + y ′ (x 0) (x − x 0) + y′′ (x0)
2!
(x − x 0) 2 + · · · .
The Taylor series, reviewed in the Example above, can be very useful to find the power
series representation of a function having infinitely many continuous derivatives.
Example 3.1.2: Find the Taylor series of y(x) = sin(x) centered at x0 = 0.
Solution: We need to compute the derivatives of the function y and evaluate these derivatives
at the point we center the expansion, in this case x 0 = 0.
y(x) = sin(x) ⇒ y(0) = 0, y ′ (x) = cos(x) ⇒ y ′ (0) = 1,
y ′′ (x) = − sin(x) ⇒ y ′′ (0) = 0, y ′′′ (x) = − cos(x) ⇒ y ′′′ (0) = −1.
Introduce the numbers y (n) (0) into Taylor’s formula,
sin(x) = x − x3
3!
+ x5
5!
− · · · ⇒ sin(x) =
∞
n=0
(−1) n
(2n + 1)! x(2n+1) .
A similar calculation leads us to the Taylor series of y(x) = cos(x) centered at x0 = 0,
∞ (−1)
cos(x) =
n
(2n)! x(2n) .
n=0
It is important to recall that the power series of a function may not be defined on the
whole domain of the function. The following example shows a function with this property.
Example 3.1.3: Find the Taylor series for y(x) = 1
1 − x centered at x0 = 0.
The power series expansion for the function y = 1
(1−x)
⊳
⊳
does not converge for |x| > 1. For
example, the power series y(x) = 1
1−x = ∞
n=0 xn evaluated at x = 1 is the sum
∞
1 n = 1 + 1 + 1 + · · · ,
n=0
1
which diverges. We conclude that the power series expansion for the function y(x) =
(1 − x)
diverges for x ∈ (−∞, −1) ∪ [1, ∞). We will show later on that this power series expansion
converges for x ∈ (−1, 1). ⊳
Later on we will need the notion of convergence of an infinite series in absolute value.

88 G. NAGY – ODE july 2, 2013
Solution: Notice that this function is well
defined for every x ∈ R − {1}. The function
graph can be seen in Fig. 12. To find
the Taylor series we need to compute the nderivative,
y (n) (0). It simple to check that,
y (n) (x) =
n!
(1 − x) n+1 , so y(n) (0) = n!.
We conclude: y(x) = 1
1 − x =
∞
x n .
n=0
Definition 3.1.2. The power series y(x) =
series
∞
|an| |x − x0| n converges.
n=0
(
−1
y
1
)
1
y ( x ) = 1 / ( 1 − x )
Figure 12. The graph of
1
y =
(1 − x) .
∞
an (x − x0) n converges absolutely iff the
n=0
∞ (−1)
Example 3.1.4: One can show that the series s =
n=1
n
converges, but this series does
n
∞ 1
not converge absolutely, since diverges. ⊳
n
n=1
Since power series representations of functions might not converge on the same domain
where the function is defined, it is useful to introduce the region where the power series
converges.
∞
Definition 3.1.3. The radius of convergence of a power series y(x) = an (x − x0) n
is the number ρ 0 that satisfies both the series converges absolutely for |x − x0| < ρ and
the series diverges for |x − x0| > ρ.
The radius of convergence defines the size of the biggest open interval where the power series
converges. This interval is symmetric around the series center point x0.
diverges
converges absolutely
( )
x
0
rho
diverges
Figure 13. Example of the radius of convergence.
Example 3.1.5: We state the radius of convergence of few power series.
1
(1) The series
1 − x =
∞
x n has radius of convergence ρ = 1.
n=0
x
n=0
x

(2) The series e x =
∞
n=0
(3) The series sin(x) =
(4) The series cos(x) =
x n
n!
∞
n=0
∞
n=0
G. NAGY – ODE July 2, 2013 89
has radius of convergence ρ = ∞.
(−1) n
(2n + 1)! x(2n+1) has radius of convergence ρ = ∞.
(−1) n
(2n)! x(2n) has radius of convergence ρ = ∞.
The next result provides a way to compute the radius of convergence of a power series.
∞
Theorem 3.1.4 (Ratio test). Given the power series y(x) = an (x − x0) n , introduce
the number L = lim
n→∞
n=0
|an+1|
. Then, the following statements hold:
|an|
(1) The power series converges in the domain |x − x0|L < 1.
(2) The power series diverges in the domain |x − x0|L > 1.
(3) The power series may or may not converge at |x − x0|L = 1.
Therefore, if L = 0, then ρ = 1
is the series radius of convergence; if L = 0, then the radius
L
of convergence is ρ = ∞.
We finish this review subsection with a comment on summation index manipulations. It
is well-known that the series
y(x) = a0 + a1(x − x0) + a2(x − x0) 2 + · · ·
is usually written using the summation notation
∞
y(x) = an (x − x0) n .
n=0
Now, the label n has nothing in particular, any other label defines the same series, for
example the labels k and m below,
∞
y(x) = ak (x − x0) k ∞
= am+3 (x − x0) m+3 .
k=0
m=−3
In the first sum we just changed the label name from n to k, that is, k = n. In the second
sum above we re-label the sum as follows, n = m + 3. Since the initial value for n is
n = 0, then the initial value of m is m = −3. Derivatives of power series can be computed
derivating every term in the power series, that is,
y ′ ∞
(x) = n an (x − x0) n−1 ∞
= n an (x − x0) n−1 = a1 + 2a2(x − x0) + · · · .
n=0
n=1
Notice that the power series for the y ′ can start either at n = 0 or n = 1, since the coefficients
have a multiplicative factor n. We will usually re-label derivatives of power series as follows,
y ′ ∞
(x) = n an (x − x0) n−1 ∞
= (m + 1) am+1 (x − x0) m
n=1
where m = n − 1, that is, n = m + 1.
m=0

90 G. NAGY – ODE july 2, 2013
3.1.2. Equations with regular points. In this subsection we concentrate in solving the
following problem: Find a power series representation of the solutions y of the variable
coefficients equation
P (x) y ′′ + Q(x) y ′ + R(x) y = 0
centered at any point x0 ∈ R where P (x0) = 0, The condition on the point x0, where we
center the power series, is so important that we give it a particular name.
Definition 3.1.5. Given continuous functions P , Q, R : (x1, x2) → R and the equation
P (x) y ′′ + Q(x) y ′ + R(x) y = 0,
a point x0 ∈ (x1, x2) is called a regular point of the equation above iff holds that P (x0) = 0.
The point x 0 is called a singular point iff holds that P (x 0) = 0.
The differential equation changes order at singular points. Since the coefficient P vanishes
at a singular point, in a neighborhood of such a point the equation is second order and first
order. This property makes finding solutions of the differential equation more difficult near
singular points than near regular points.
Example 3.1.6: The differential equation given in (3.1.3) has a singular point at x0 = 2,
since the function P (x) = (x − 2) 2 vanishes at that point. Any other point is a regular
point. ⊳
Here is a summary of the power series method to find solutions to differential equations
of the form given in Eq. (3.1.1) near regular points x0.
(a) Verify that the power series expansion is centered at a regular point, x 0;
(b) Propose a power series representation of the solution centered at x0, given by
∞
y(x) = an (x − x0) n ;
n=0
(c) Introduce the power series expansion above into the differential equation and find a
recurrence relation among the coefficients an;
(d) Solve the recurrence relation in terms of free coefficients;
(e) If possible, add up the resulting power series for the solution function y.
We follow these steps in the examples below to find solutions to several differential equations.
We start with a first order constant coefficient equation, and then we continue with
a second order constant coefficient equation. The last two examples consider variable coefficient
equations.
Example 3.1.7: Find a power series solution y around the point x 0 = 0 of the equation
y ′ + c y = 0, c ∈ R.
Solution: We already know every solution to this equation. This is a first order, linear,
differential equation, so using the method of integrating factor we find that the solution is
y(x) = a 0 e −c x , a 0 ∈ R.
We are now interested in obtaining such solution with the power series method. Although
this is not a second order equation, the power series method still works in this example.
Propose a solution of the form
∞
y = an x n ⇒ y ′ ∞
= nan x (n−1) .
n=0
n=1

G. NAGY – ODE July 2, 2013 91
We can start the sum in y ′ at n = 0 or n = 1. We choose n = 1, since it is more convenient
later on. Introduce the expressions above into the differential equation,
∞
nan x n−1 ∞
+ c an x n = 0.
n=1
Re-label the first sum above so that the functions x n−1 and x n in the first and second sum
have the same label. One way is the following,
∞
n=0
n=0
(n + 1)a (n+1) x n +
∞
c an x n = 0
n=0
We can now write down both sums into one single sum,
∞ n
(n + 1)a(n+1) + c an x = 0.
n=0
Since the function on the left-hand side must be zero for every x ∈ R, we conclude that
every coefficient that multiplies x n must vanish, that is,
(n + 1)a (n+1) + c an = 0, n 0.
The last equation is called a recurrence relation among the coefficients an. The solution of
this relation can be found by writing down the first few cases and then guessing the general
expression for the solution, that is,
n = 0, a1 = −c a0 ⇒ a1 = −c a0,
n = 1, 2a 2 = −c a 1 ⇒ a 2 = c2
2! a 0,
n = 2, 3a3 = −c a2 ⇒ a3 = − c3
3! a0,
n = 3, 4a4 = −c a3 ⇒ a4 = c4
4! a0.
One can check that the coefficient an can be written as
n cn
an = (−1)
n! a0,
which implies that the solution of the differential equation is given by
∞
n cn
y(x) = a0 (−1)
n! xn ∞ (−c x)
⇒ y(x) = a0 n
n!
n=0
n=0
⇒ y(x) = a 0 e −c x .
Example 3.1.8: Find a power series solution y(x) around the point x 0 = 0 of the equation
y ′′ + y = 0.
Solution: We know that the solution can be found computing the roots of the characteristic
polynomial r 2 + 1 = 0, which gives us the solutions
y(x) = a 0 cos(x) + a 1 sin(x).
We now recover this solution using the power series,
∞
∞
y =
n=0
an x n ⇒ y ′ =
n=1
nan x (n−1) , ⇒ y ′′ =
∞
n(n − 1)an x (n−2) .
n=2
⊳

92 G. NAGY – ODE july 2, 2013
Introduce the expressions above into the differential equation, which involves only the function
and its second derivative,
∞
n(n − 1)an x n−2 ∞
+ an x n = 0.
n=2
Re-label the first sum above, so that both sums have the same factor xn . One way is,
∞
(n + 2)(n + 1)a (n+2) x n ∞
+ an x n = 0.
n=0
Now we can write both sums using one single sum as follows,
∞
n
x = 0 ⇒ (n + 2)(n + 1)a(n+2) + an = 0. n 0.
n=0
(n + 2)(n + 1)a(n+2) + an
The last equation is the recurrence relation. The solution of this relation can again be found
by writing down the first few cases, and we start with even values of n, that is,
n = 0, (2)(1)a2 = −a0 ⇒ a2 = − 1
2! a0,
n=0
n=0
n = 2, (4)(3)a4 = −a2 ⇒ a4 = 1
4! a0,
n = 4, (6)(5)a6 = −a4 ⇒ a6 = − 1
6! a0.
One can check that the even coefficients a2k can be written as
a2k = (−1)k
(2k)! a0.
The coefficients an for the odd values of n can be found in the same way, that is,
n = 1, (3)(2)a3 = −a1 ⇒ a3 = − 1
3! a1,
n = 3, (5)(4)a5 = −a3 ⇒ a5 = 1
5! a1,
n = 5, (7)(6)a7 = −a5 ⇒ a7 = − 1
7! a1.
One can check that the odd coefficients a2k+1 can be written as
a2k+1 = (−1)k
(2k + 1)! a1.
Split the sum in the expression for y into even and odd sums. We have the expression for
the even and odd coefficients. Therefore, the solution of the differential equation is given by
y(x) = a 0
∞
k=0
(−1) k
(2k)! x2k + a 1
∞
k=0
(−1) k
(2k + 1)! x2k+1 .
One can check that these are precisely the power series representations of the cosine and
sine functions, respectively,
y(x) = a0 cos(x) + a1 sin(x).
⊳

G. NAGY – ODE July 2, 2013 93
Example 3.1.9: Find the first four terms of the power series expansion around the point
x0 = 1 of each fundamental solution to the differential equation
y ′′ − x y ′ − y = 0.
Solution: This is a differential equation we cannot solve with the methods of previous
sections. This is a second order, variable coefficients equation. We use the power series
method, so we look for solutions of the form
∞
y = an(x − 1) n ⇒ y ′ ∞
= nan(x − 1) n−1 ⇒ y ′′ ∞
= n(n − 1)an(x − 1) n−2 .
n=0
n=1
We start working in the middle term in the differential equation. Since the power series is
centered at x 0 = 1, it is convenient to re-write this term as x y ′ = [(x − 1) + 1] y ′ , that is,
x y ′ =
=
∞
nanx(x − 1) n−1
n=1
∞
n=1
n=1
n−1
nan (x − 1) + 1 (x − 1)
n=2
∞
= nan(x − 1) n ∞
+ nan(x − 1) n−1 . (3.1.4)
As usual by now, the first sum on the right-hand side of Eq. (3.1.4) can start at n = 0, since
we are only adding a zero term to the sum, that is,
∞
nan(x − 1) n ∞
= nan(x − 1) n ;
n=1
while it is convenient to re-label the second sum in Eq. (3.1.4) follows,
∞
nan(x − 1) n−1 ∞
= (n + 1)a (n+1)(x − 1) n ;
n=1
so both sums in Eq. (3.1.4) have the same factors (x − 1) n . We obtain the expression
x y ′ ∞
= nan(x − 1) n ∞
+ (n + 1)a (n+1)(x − 1) n
=
n=0
n=0
n=0
n=0
n=1
∞ n
nan + (n + 1)a (n+1) (x − 1) . (3.1.5)
n=0
In a similar way re-label the index in the expression for y ′′ , so we obtain
y ′′ ∞
=
n=0
(n + 2)(n + 1)a (n+2)(x − 1) n . (3.1.6)
If we use Eqs. (3.1.5)-(3.1.6) in the differential equation, together with the expression for y,
the differential equation can be written as follows
∞
(n + 2)(n + 1)a (n+2)(x − 1) n ∞
∞
n
− nan + (n + 1)a (n+1) (x − 1) − an(x − 1) n = 0.
n=0
n=0
We can now put all the terms above into a single sum,
∞
n=0
(n + 2)(n + 1)a (n+2) − (n + 1)a (n+1) − nan − an
n=0
(x − 1) n = 0.

94 G. NAGY – ODE july 2, 2013
This expression provides the recurrence relation for the coefficients an with n 0, that is,
which can be rewritten as follows,
(n + 2)(n + 1)a (n+2) − (n + 1)a (n+1) − (n + 1)an = 0
(n + 1)
= 0,
(n + 2)a (n+2) − a (n+1) − an
(n + 2)a (n+2) − a (n+1) − an = 0. (3.1.7)
We can solve this recurrence relation for the first four coefficients,
n = 0 2a 2 − a 1 − a 0 = 0 ⇒ a 2 = a 1
2 + a 0
2 ,
n = 1 3a 3 − a 2 − a 1 = 0 ⇒ a 3 = a1
2
+ a0
6 ,
n = 2 4a4 − a3 − a2 = 0 ⇒ a4 = a1 a0
+
4 6 .
Therefore, the first terms in the power series expression for the solution y of the differential
equation are given by
a0
y = a0 + a1(x − 1) +
2 + a
1
(x − 1)
2
2
a0
+
6 + a
1
(x − 1)
2
3
a0
+
6 + a
1
(x − 1)
4
4 + · · ·
which can be rewritten as
y = a0 1 + 1
2 (x − 1)2 + 1
6 (x − 1)3 + 1
6 (x − 1)4
+ · · ·
+ a1 (x − 1) + 1
2 (x − 1)2 + 1
2 (x − 1)3 + 1
4 (x − 1)4
+ · · ·
So the first four terms on each fundamental solution are given by
y 1 = 1 + 1
2 (x − 1)2 + 1
6 (x − 1)3 + 1
6 (x − 1)4 ,
y 2 = (x − 1) + 1
2 (x − 1)2 + 1
2 (x − 1)3 + 1
4 (x − 1)4 .
Example 3.1.10: Find the first three terms of the power series expansion around the point
x0 = 2 of each fundamental solution to the differential equation
y ′′ − x y = 0.
Solution: We then look for solutions of the form
∞
y = an(x − 2) n .
n=0
It is convenient to rewrite the function x y = [(x − 2) + 2] y, that is,
∞
xy = anx(x − 2) n
=
n=0
∞
n=0
n=0
n
an (x − 2) + 2 (x − 2)
∞
= an(x − 2) n+1 ∞
+ 2an(x − 2) n . (3.1.8)
n=0
⊳

G. NAGY – ODE July 2, 2013 95
We now relabel the first sum on the right-hand side of Eq. (3.1.8) in the following way,
∞
an(x − 2) n+1 ∞
=
n=0
n=1
We do the same type of re-labeling on the expression for y ′′ ,
y ′′ ∞
= n(n − 1)an(x − 2) n−2
=
n=2
∞
n=0
(n + 2)(n + 1)a (n+2)(x − 2) n .
Then, the differential equation above can be written as follows
∞
(n + 2)(n + 1)a (n+2)(x − 2) n ∞
− 2an(x − 2) n ∞
−
n=0
(2)(1)a2 − 2a0 +
n=0
a (n−1)(x − 2) n . (3.1.9)
n=1
a (n−1)(x − 2) n = 0
∞
(n + 2)(n + 1)a (n+2) − 2an − a (n−1) (x − 2) n = 0.
n=1
So the recurrence relation for the coefficients an is given by
a2 − a0 = 0, (n + 2)(n + 1)a (n+2) − 2an − a (n−1) = 0, n 1.
We can solve this recurrence relation for the first four coefficients,
n = 0 a2 − a0 = 0 ⇒ a2 = a0,
n = 1 (3)(2)a3 − 2a1 − a0 = 0 ⇒ a3 = a0
6
+ a1
3 ,
n = 2 (4)(3)a4 − 2a2 − a1 = 0 ⇒ a4 = a0 a1
+
6 12 .
Therefore, the first terms in the power series expression for the solution y of the differential
equation are given by
y = a0 + a1(x − 2) + a0(x − 2) 2
a0 a1
+ + (x − 2)
6 3
3
a0 a1
+ + (x − 2)
6 12
4 + · · ·
which can be rewritten as
y = a0 1 + (x − 2) 2 + 1
6 (x − 2)3 + 1
6 (x − 2)4
+ · · ·
+ a1 (x − 2) + 1
3 (x − 2)3 + 1
12 (x − 2)4
+ · · ·
So the first three terms on each fundamental solution are given by
y 1 = 1 + (x − 2) 2 + 1
6 (x − 2)3 ,
y 2 = (x − 2) + 1
3 (x − 2)3 + 1
12 (x − 2)4 .
⊳

96 G. NAGY – ODE july 2, 2013
3.1.3. Exercises.
3.1.1.- . 3.1.2.- .

G. NAGY – ODE July 2, 2013 97
3.2. The Euler equation
3.2.1. The main result. In this Section we study how to find solutions of certain differential
equations with singular points, for example an equation like this one:
(x − 2) 2 y ′′ + 6(x − 2) y ′ + 6 y = 0.
We are interested in finding the solution values y(x) for x arbitrary close to the singular
point x0 = 2. Therefore, we are not interested in a power series expansion of the solution
around a regular point, which in the example above is any point different from x0 = 2. In
the previous Section we have stated without proof that the solutions defined arbitrary close
to x0 = 2 are given by
1
y1(x) =
(x − 2) 2 , y 1
2(x) = .
(x − 2) 3
In this Section we show how to obtain these solutions. We start by introducing the type of
equations we are interested in.
Definition 3.2.1. Given real constants p0, q0, the Euler differential equation for the
unknown y with singular point at x0 ∈ R is given by
(x − x0) 2 y ′′ + p0 (x − x0) y ′ + q0 y = 0.
We take the time to raise few comments of this Euler differential equation. First, this is
a variable coefficients equation. So, it is clear that the exponential functions y(x) = e rx are
not solutions of the Euler equation. Just introduce such a function into the equation, and
it is simple to show that there is no r such that the exponential is solution. Second, notice
that the point x0 ∈ R is a singular point of the equation. The particular case x0 = 0 is
x 2 y ′′ + p0 x y ′ + q0 y = 0.
The following result summarizes what is known about solutions of an Euler equation.
Theorem 3.2.2 (Euler equation). Given p0, q0, and x0 ∈ R consider the Euler equation
(x − x0) 2 y ′′ + p0 (x − x0) y ′ + q0 y = 0. (3.2.1)
Let r± be the roots of the Euler characteristic polynomial p(r) = r(r − 1) + p 0r + q 0.
(a) If r+ = r−, then the general solution of Eq. (3.2.1) is given by
y(t) = c0|x − x0| r- + c1|x − x0| r- , x = x0, c0, c1 ∈ R.
(b) If r+ = r− ∈ R, then the general solution of Eq. (3.2.1) is given by
y(t) = c0 + c1 ln |x − x0|
|x − x0| r- , x = x0, c0, c1 ∈ R.
Furthermore, given real constants x 1 = x 0, y 0 and y 1, there is a unique solution to the initial
value problem given by Eq. (3.2.1) and the initial conditions
y(x 1) = y 0, y ′ (x 1) = y 1.
Remark: Although we call the solutions found in (a)-(b) the general solution, they do not
include all possible solutions to the differential equation. For example, consider the case that
the roots of the Euler characteristic polynomial are different and one root is odd, say r+ is
odd and r+ = r−. Then, the formula provided in (a) is given by y = c1|x−x0| r- +c2|x−x0| r- ,
but this expression does not include the solution y = (x − x0) r- . So, to simplify the writing
of the Theorem above we do not include all possible solutions to the equation, although we
keep the name general solution for the solutions we present in that Theorem.

98 G. NAGY – ODE july 2, 2013
We do not provide a proof of this Theorem. We just mention the main idea behind this
proof. To understand this main idea it is useful to recall what we did to find solutions to
constant coefficient equations of the form
y ′′ + a1 y ′ + a0 y = 0.
Since the derivatives of the exponential function are proportional to the same exponential,
that is, e rx ′ = re rx , one can see that introducing this exponential y = e rx into the equation
above converts the differential equation for y into an algebraic equation for r as follows,
(r 2 + a1r + a0) e rx = 0 ⇔ r 2 + a1r + a0 = 0.
This is the central idea to find solutions to a constant coefficient equation, that the exponential
function disappears from the equation, leaving only an algebraic equation for the
coefficient r. Later on one can show that the solutions obtained in this way are in fact all
possible solution, as it was shown in the proof of Theorem 2.2.2.
We have seen that exponential functions are not appropriate to find solutions of the Euler
equation. But the idea above can be carried out on the Euler equation
(x − x0) 2 y ′′ + p0 (x − x0) y ′ + q0 y = 0,
we just need a different function. Now, recall that power functions y = (x − x0) r satisfy
y ′ = r(x − x0) r−1 ⇒ (x − x0)y ′ = r(x − x0) r .
A similar property holds for the second derivative,
y ′′ = r(r − 1)(x − x0) r−2 ⇒ (x − x0) 2 y ′ = r(r − 1)(x − x0) r .
Hence, introducing this power function into the Euler equation above transform this differential
equation for y into an algebraic equation for r as follows
r(r − 1) + p0r + q0 (x − x0) r = 0 ⇔ r(r − 1) + p0r + q0 = 0.
The constant r must be a root of the Euler characteristic polynomial p(r) = r(r−1)+p0r+q0.
And we again have different cases depending whether this polynomial has two different roots
or only one. These cases are described in Theorem 3.2.2. Notice that this Theorem asserts
more than that. The Theorem asserts that these are the only solutions of the Euler equation,
and we are not proving this here.
Example 3.2.1: Find the general solution of the Euler equation
x 2 y ′′ + 4x y ′ + 2 y = 0.
Solution: We look for solutions of the form y(x) = x r , which implies that
x y ′ (x) = rx r , x 2 y ′′ (x) = r(r − 1) x r ,
therefore, introducing this function y into the differential equation we obtain
r(r − 1) + 4r + 2 x r = 0 ⇔ r(r − 1) + 4r + 2 = 0.
The solutions are computed in the usual way,
r 2 + 3r + 2 = 0 ⇒ r = 1
√
−3 ± 9 − 8 ⇒
2
So the general solution of the differential equation above is given by
y(x) = c1 x −1 + c2 x −2 .
r1 = −1
r2 = −2.
⊳

G. NAGY – ODE July 2, 2013 99
Example 3.2.2: Find the general solution of the Euler equation
x 2 y ′′ − 3x y ′ + 13 y = 0.
Solution: We look for solutions of the form y(x) = x r , which implies that
x y ′ (x) = rx r , x 2 y ′′ (x) = r(r − 1) x r ,
therefore, introducing this function y into the differential equation we obtain
r(r − 1) − 3r + 13 x r = 0 ⇔ r(r − 1) − 3r + 13 = 0.
The solutions are computed in the usual way,
r 2 − 4r + 13 = 0 ⇒ r = 1
√
4 ± −36 ⇒
2
So the general solution of the differential equation above is given by
r1 = 2 + 3i
r2 = 2 − 3i.
y(x) = c1 x (2+3i) + c2 x (2−3i) . (3.2.2)
⊳
3.2.2. The complex-valued roots. In the case that the roots of the Euler characteristic
polynomial are complex-valued, there always is linear combination of complex-valued fundamental
solutions that results in real-valued fundamental solutions. The reason is that the
Euler differential equation coefficients P 0, q 0 are real-valued. For example, the solution of
the example above, given in Eq. (3.2.2) can also be expressed as follows,
y(t) = ˜c1 x 2 cos ln(|x| 3 ) + ˜c2 x 2 sin ln(|x| 3 ) , (3.2.3)
where the constants ˜c 1, ˜c 2 are different form the constants c1, c2 present in (3.2.2). The
steps needed to obtain Eq. (3.2.3) are summarized in the proof of the Corollary below.
Corollary 3.2.3 (Complex roots). If p0, q0 ∈ R satisfy that (p0 − 1) 2
− 4q0 < 0, then
the indicial polynomial p(r) = r(r − 1) + p0r + q0 of the Euler equation
has complex roots r± = α ± iβ, where
(x − x0) 2 y ′′ + p0(x − x0) y ′ + q0 y = 0 (3.2.4)
4q0 − (p 0 − 1) 2 .
α = − (p0 − 1)
, β =
2
1
2
Furthermore, a fundamental set of solution to Eq. (3.2.4) is
˜y 1(x) = |x − x 0| (α+iβ) , ˜y 2(x) = |x − x 0| (α−iβ) ,
while another fundamental set of solutions to Eq. (3.2.4) is
y1(x) = |x − x0| α cos β ln |x − x0| , y2(x) = |x − x0| α sin β ln |x − x0| .
Proof of Corollary 3.2.3: For simplicity consider the case x0 = 0. Since the functions
˜y1(x) = x (α+iβ) , ˜y2(x) = x (α−iβ) ,
are solutions to the differential equation in the Corollary, then so are the functions
y1(x) = 1
˜y1(x) + ˜y2(x)
2
, y2(x) = 1
˜y1(x) − ˜y2(x)
2i
.
Now, we need to recall several relations. First, recall the main property of exponentials,
e (α±iβ)t = e αt e ±iβt . Second, recall the definition of the power function for complex exponents,
that is,
x iβ = e ln(|x|iβ ) = e iβ ln(|x|) .

100 G. NAGY – ODE july 2, 2013
Third, recall the Euler equation for complex exponentials,
e iβt = cos(βt) + i sin(βt),
e −iβt = cos(βt) − i sin(βt).
All these properties above are used in the following calculation:
y1(x) = 1
2 xα x iβ + x −iβ
= 1
2 xαe
ln(|x|iβ ) ln(|x|
+ e −iβ )
e iβ ln(|x|) −iβ
+ e
ln(|x|)
= 1
2 xα
= x α cos β ln(|x|)
= x α cos ln(|x| β ) ,
We have then concluded that
y2(x) = 1
2i xα x iβ − x −iβ
= 1
2i xαe
ln(|x|iβ ) ln(|x|
− e −iβ )
e iβ ln(|x|) −iβ
− e
ln(|x|)
= 1
2i xα
= x α sin β ln(|x|)
= x α sin ln(|x| β ) ,
y1(x) = x α cos ln(|x| β ) , y2(x) = x α sin ln(|x| β ) .
This establishes the Corollary.
Example 3.2.3: Find a real-valued general solution of the Euler equation
x 2 y ′′ − 3x y ′ + 13 y = 0.
Solution: The indicial equation is r(r − 1) − 3r + 13 = 0, with solutions
A complex-valued general solution is
A real-valued general solution is
r 2 − 4r + 13 = 0 ⇒ r+ = 2 + 3i, r− = 2 − 3i.
y(x) = ˜c1 |x| (2+3i) + ˜c2 |x| (2−3i)
˜c1, ˜c2 ∈ C.
y(x) = c1 |x| 2 cos 3 ln |x| + c2 |x| 2 sin 3 ln |x| , c1, c2 ∈ R.
Our last example considers the case where the Euler characteristic polynomial has a
repeated root.
Example 3.2.4: Find the general solution of the Euler equation
x 2 y ′′ − 3x y ′ + 4 y = 0.
Solution: We look for solutions of the form y(x) = x r , then the constant r must be solution
of the Euler characteristic polynomial
r(r − 1) − 3r + 4 = 0 ⇔ r 2 − 4r + 4 = 0 ⇒ r+ = r− = 2.
Therefore, the general solution of the Euler equation in this case is given by
y(x) = c1x 2 + c2x 2 ln(|x|).
⊳
⊳

3.2.3. Exercises.
3.2.1.- . 3.2.2.- .
G. NAGY – ODE July 2, 2013 101

102 G. NAGY – ODE july 2, 2013
3.3. Regular-singular points
In Sect. 3.2 we have introduced the Euler equation and we have seen that contains one
singular point. In the same Section we have seen how to find solutions to this Euler equation
that are well-defined arbitrary close to the singular point. In this Section we introduce a
large set of differential equations containing singular points similar to the singular points
of the Euler equation. We will call them differential equations with regular-singular points,
where the name tries to suggest that they are singular points which happen to be not
so singular. This type of equations include the equations with regular points studied in
Section 3.1 and the Euler equation studied in Section 3.2. We show how to find at least
one solution well-defined arbitrary close to the regular-singular points to this broad type of
equations.
3.3.1. Finding regular-singular points. Recall that that a point x0 ∈ R is a singular
point of the second order, linear, differential equation
P (x) y ′′ + Q(x) y ′ + R(x) y = 0
iff holds that P (x0) = 0. Regular-singular points are singular points which are not so
singular. Here is a precise definition.
Definition 3.3.1. A singular point x0 ∈ R of the equation
P (x) y ′′ + Q(x) y ′ + R(x) y = 0
is called a regular-singular point iff the following limits are finite,
and both functions
lim
x→x0
(x − x0) Q(x)
, lim
P (x)
x→x0
(x − x0) Q(x)
,
P (x)
admit convergent Taylor series expansions around x0.
(x − x0) 2 R(x)
,
P (x)
(x − x0) 2 R(x)
,
P (x)
Example 3.3.1: Show that the singular point of every Euler equation is a regular-singular.
Solution: Given real constants p0, q0, x0, consider the general Euler equation
(x − x0) 2 y ′′ + p0(x − x0) y ′ + q0 y = 0.
In this case the functions P , Q and R introduced in the Definition 3.3.1 are given by
Therefore, we obtain,
P (x) = (x − x0) 2 , Q(x) = p0(x − x0), R(x) = q0.
(x − x0) Q(x)
P (x)
= p0,
(x − x0) 2 R(x)
P (x)
= q0.
From these equations is trivial to see that both conditions given in the Definition 3.3.1 are
satisfied by the coefficients of every Euler equation. Therefore, the singular point of every
Euler equation is regular-singular. ⊳
Example 3.3.2: Find the regular-singular points of the differential equation
where α is a real constant.
(1 − x 2 ) y ′′ − 2x y ′ + α(α + 1) y = 0,

G. NAGY – ODE July 2, 2013 103
Solution: We first find the singular points of this equation, which are given by the zeros
of the coefficient function that multiplies y ′′ , that is,
P (x) = (1 − x 2
x0 = 1,
) = (1 − x)(1 + x) = 0 ⇒
x1 = −1.
Let us start with the singular point x0 = 1. We then have
(x − 1) Q(x)
P (x)
= (x − 1)(−2x)
(1 − x)(1 + x)
= 2x
1 + x ,
(x − 1) 2 R(x)
P (x)
= (x − 1)2α(α + 1)
(1 − x)(1 + x)
= (x − 1)α(α + 1)
;
1 + x
which implies that both functions above have power series expansions around x0 = 1.
Furthermore, the following limits are finite, namely,
(x − 1) Q(x)
(x − 1)
lim
= 1; lim
x→1 P (x)
x→1
2 R(x)
= 0.
P (x)
We conclude that x 0 = 1 is a regular-singular point of the differential equation. We now
perform the same calculation with the point x 1 = −1, that means,
(x + 1) Q(x)
P (x)
= (x + 1)(−2x)
(1 − x)(1 + x)
= − 2x
1 − x ,
(x + 1) 2 R(x)
P (x)
= (x + 1)2 α(α + 1)
(1 − x)(1 + x)
= (x + 1)α(α + 1)
;
1 − x
which implies that both functions above have power series expansions around x1 = −1.
Furthermore, the following limits are finite, namely,
(x + 1) Q(x)
(x + 1)
lim
= 1; lim
x→−1 P (x)
x→−1
2 R(x)
= 0.
P (x)
Therefore, the point x1 = −1 is also a regular-singular point of the differential equation. ⊳
Example 3.3.3: Find the regular-singular points of the differential equation
(x + 2) 2 (x − 1) y ′′ + 3(x − 1) y ′ + 2 y = 0.
Solution: The singular point of the differential equation above are given by x0 = −2 and
x 1 = 1. In the first case we have that
(x + 2)Q(x) (x + 2)3(x − 1)
lim
= lim
x→−2 P (x) x→−2 (x + 2) 2 = lim
(x − 1) x→−2
3
= ∞,
(x + 2)
so x0 = −2 is not a regular-singular point. In the second case, x1 = 1, we obtain
(x − 1) Q(x)
P (x)
= (x − 1)2(x − 1)
(x + 2)(x − 1)
2(x − 1)
= − ,
(x + 2) 2
(x − 1) 2 R(x)
P (x)
2(x − 1)
=
2
(x + 2) 2 (x − 1)
= 2(x − 1)
;
(x + 2) 2
which implies that both functions above have power series expansions around x1 = 1.
Furthermore, the following limits are finite, namely,
(x − 1) Q(x)
(x − 1)
lim
= 0; lim
x→1 P (x)
x→1
2 R(x)
= 0.
P (x)
Therefore, the point x1 = −1 is a regular-singular point of the differential equation. ⊳

104 G. NAGY – ODE july 2, 2013
Example 3.3.4: Find the regular-singular points of the differential equation
x y ′′ − x ln(|x|) y ′ + 3x y = 0.
Solution: The singular point is x 0 = 0. Although the following limits are finite,
xQ(x) x
lim = lim
x→0 P (x) x→0
x ln(|x|)
ln(|x|)
= lim
x
x→0 1 = lim
x→0
x
x
lim
x→0
2R(x) P (x)
x
= lim
x→0
2 (3x)
x
1
x
− 1
x 2
= lim
x→0 3x2 = 0,
= lim
x→0 −x = 0,
the point x0 = 0 is not a regular-singular point, since the function xQ/P does not have a
power series expansion around zero, since
xQ(x)
P (x)
= x ln(|x|),
and the log function does not have a Taylor series at x0 = 0. ⊳
3.3.2. Solving equations with regular-singular points. In the last part of this Section
we study how to find solutions to differential equations having regular-singular points. We
are interested in solutions well-defined arbitrary close to the regular-singular point. The
main idea is this: Since near a regular-singular point the coefficients of the differential
equation are close to the coefficients of a particular Euler equation, then it could happen
that near that regular-singular point their solutions could be also close to each other. More
precisely, if x0 is a regular-singular point of
with limits
P (x) y ′′ + Q(x) y ′ + R(x) y = 0,
(x − x0)Q(x) (x − x0) lim
= p0 and lim
x→x0 P (x)
x→x0
2R(x) = q0, P (x)
then the coefficients of the differential equation above near x0 are close to the coefficients
of the Euler equation
(x − x0) 2 y ′′ + p0(x − x0) y ′ + q0 y = 0.
Since the differential equation is close to an Euler equation, then it could happen that the
solutions of the differential equation might be close to the solutions of an Euler equation.
And we know that at least one solution of an Euler equation is
ye(x) = (x − x 0) r , (3.3.1)
where the constant r is the root of the Euler characteristic polynomial, and x0 is the regularsingular
point of the original equation. Therefore, the idea is to look for solutions of the
original differential equation having the form
y(x) = (x − x0) r
∞
an (x − x0) n ∞
⇔ y(x) = an (x − x0) (n+r) . (3.3.2)
n=0
It is simple to see that this expression of the solution contains the idea that the values y(x)
are close to ye(x) for values of x close to x0, since in this latter case we have that
n=0
y(x) = ye(x) a0 + a1(x − x0) + · · · a0 ye(x) for x x0,
where the symbol a b, with a, b ∈ R means that |a − b| is close to zero.
We remark that it could be possible that not every solution of a differential equation
containing regular-singular points will have the form given in Eq. (3.3.2). However, it can
be shown that at least one solution will have this form. A similar situation occurs with the

G. NAGY – ODE July 2, 2013 105
solutions to the Euler equation, where there are cases with solutions containing logarithmic
terms, which means that they do not have the form given in Eq. (3.3.1). For an explicit
example of this situation, see the comment at the end of Example 3.3.5.
A summary of the method to find solutions to differential equations having regularsingular
points is the following:
∞
(1) Look for a solution y of the form y(x) =
n=0
an (x − x0) (n+r) ;
(2) Introduce this power series expansion into the differential equation and find both the
exponent r and a recurrence relation for the coefficients an;
(3) First find the solutions for the constant r. Then, introduce this result for r into the
recurrence relation for the coefficients an. Only then, solve this latter recurrence relation
for the coefficients an.
Example 3.3.5: Find the solution y near the regular-singular point x0 = 0 of the equation
x 2 y ′′ − x(x + 3) y ′ + (x + 3) y = 0.
Solution: We propose a solution of the form
∞
y(x) = an x (n+r) .
n=0
The first and second derivatives are given by
y ′ ∞
(x) =
n=0
(n + r)an x (n+r−1) , y ′′ (x) =
∞
(n + r)(n + r − 1)an x (n+r−2) .
In the case r = 0 we had the relation ∞ n=0 nan x (n−1) = ∞ n=1 nan x (n−1) . But in our
case r = 0, so we do not have the freedom to change in this way the starting value of the
summation index n. If we want to change the initial value for n, we have to re-label the
summation index. We now introduce these expressions into the differential equation. It is
convenient to do this step by step. We start with the term (x + 3)y, which has the form,
∞
(x + 3) y = (x + 3) an x (n+r)
=
∞
n=0
n=1
n=0
an x (n+r+1) +
n=0
∞
3an x (n+r)
n=0
∞
= a (n−1) x (n+r) ∞
+ 3an x (n+r) . (3.3.3)
n=0
We continue with the term containing y ′ , that is,
−x(x + 3) y ′ = −(x 2 ∞
+ 3x) (n + r)an x (n+r−1)
n=0
n=0
∞
= − (n + r)an x (n+r+1) ∞
− 3(n + r)an x (n+r)
n=1
n=0
∞
= − (n + r − 1)a (n−1) x (n+r) ∞
− 3(n + r)an x (n+r) . (3.3.4)
n=0

106 G. NAGY – ODE july 2, 2013
Then, we compute the term containing y ′′ as follows,
x 2 y ′′ = x 2
=
∞
(n + r)(n + r − 1)an x (n+r−2)
n=0
∞
(n + r)(n + r − 1)an x (n+r) . (3.3.5)
n=0
As one can see from Eqs.(3.3.3)-(3.3.5), the guiding principle to rewrite each term is to
have the power function x (n+r) labeled in the same way on every term. For example, in
Eqs.(3.3.3)-(3.3.5) we do not have a sum involving terms with factors x (n+r−1) or factors
x (n+r+1) . Then, the differential equation can be written as follows,
∞
(n + r)(n + r − 1)an x (n+r) ∞
− (n + r − 1)a (n−1) x (n+r)
n=0
n=0
n=1
n=1
∞
− 3(n + r)an x (n+r) ∞
+ a (n−1) x (n+r) ∞
+ 3an x (n+r) = 0.
In the equation above we need to split the sums containing terms with n 0 into the term
n = 0 and a sum containing the terms with n 1, that is,
r(r − 1) − 3r + 3 a0x r +
∞
(n + r)(n + r − 1)an − (n + r − 1)a (n−1) − 3(n + r)an + a (n−1) + 3an x (n+r) = 0,
n=1
and this expression can be rewritten as follows,
r(r − 1) − 3r + 3 a0x r +
∞ (n
+ r)(n + r − 1) − 3(n + r) + 3 an − (n + r − 1 − 1)a (n−1) x (n+r) = 0
n=1
and then,
r(r − 1) − 3r + 3 a0x r +
∞ (n
+ r)(n + r − 1) − 3(n + r − 1) an − (n + r − 2)a (n−1) x (n+r) = 0
hence,
n=1
r(r − 1) − 3r + 3 a0x r +
n=0
∞
(n + r − 1)(n + r − 3)an − (n + r − 2)a (n−1) x (n+r) = 0.
n=1
The Euler characteristic equation and the recurrence relation are given by the equations
r(r − 1) − 3r + 3 = 0, (3.3.6)
(n + r − 1)(n + r − 3)an − (n + r − 2)a (n−1) = 0. (3.3.7)
The way to solve these equations in (3.3.6)-(3.3.7) is the following: First, solve Eq. (3.3.6) for
the exponent r, which in this case has two solutions r±; second, introduce the first solution
r+ into the recurrence relation in Eq. (3.3.7) and solve for the coefficients an; the result is a
solution y+ of the original differential equation; then introduce the second solution r− into

G. NAGY – ODE July 2, 2013 107
Eq. (3.3.7) and solve again for the coefficients an; the new result is a second solution y−.
Let us follow this procedure in the case of the equations above:
r 2 − 4r + 3 = 0 ⇒ r± = 1
√ r+ = 3,
4 ± 16 − 12 ⇒
2
r− = 1.
Introducing the value r+ = 3 into Eq. (3.3.7) we obtain
(n + 2)n an − (n + 1)an−1 = 0.
One can check that the solution y+ obtained form this recurrence relation is given by
y+(x) = a0 x 3
1 + 2 1
x +
3 4 x2 + 1
15 x3
+ · · · .
Introducing the value r− = 1 into Eq. (3.3.7) we obtain
n(n − 2)an − (n − 1)an−1 = 0.
One can also check that the solution y− obtained form this recurrence relation is given by
y−(x) = a2 x x 2 + 2
3 x3 + 1
4 x4 + 1
15 x5
+ · · · ,
= a2 x 3
1 + 2 1
x +
3 4 x2 + 1
15 x3
+ · · · ⇒ y− = a2
y+.
a1
Remark: We have found that the solution y− is not linearly independent of the solution
y+. This Example shows an important property of the method described in this Section
to find solutions of equations near a regular-singular point; a property that we already
mentioned earlier in this Section: The method does not provide all solutions of a differential
equation near a regular-singular point, it only provides at least one solution near a regularsingular
point. It can be shown the following result: If the roots of the Euler characteristic
polynomial r+, r− differ by an integer, then the corresponding solutions y+, y− are linearly
dependent. It can be proven that second solution linearly independent to y+ exists, but it
has the solution a more complicated expression than the one given in Eq. (3.3.2). This new
solution involves logarithmic terms. We do not study such solutions in these notes. ⊳

108 G. NAGY – ODE july 2, 2013
3.3.3. Exercises.
3.3.1.- . 3.3.2.- .

G. NAGY – ODE July 2, 2013 109
Chapter 4. The Laplace Transform
In this Chapter we introduce and develop an idea to find solutions to differential equations
that is radically different from all we have studied so far. The idea is centered around the
Laplace Transform of a function. The Laplace Transform is a functional, that is, is a map
that transforms a function into a new function. The Laplace Transform converts derivatives
into multiplications, hence it converts differential equations into algebraic equations.
For certain types of functions the Laplace Transform is invertible. So, Laplace transforms
can be used to solve differential equations: First transform the whole differential equation
into an algebraic equation; second solve the algebraic equation; third transform back the
result, which is the solution to the differential equation. Laplace Transforms can be used
to find solutions to differential equations with arbitrary source functions but with constant
coefficients.
4.1. Definition and properties
The Laplace Transform is defined by a particular improper integral, so we start this
Section with a brief review on improper integral and few examples. Then we introduce the
definition of the Laplace Transform and then present several examples. In the following
Sections we explain how to use such transform to find solutions to differential equations.
The Laplace transform method is useful to solve non-homogeneous differential equations,
but only with constant coefficients.
4.1.1. Review of improper integrals. Improper integrals are integrals on unbounded
domains. They are defined as a limit of definite integrals. More precisely,
∞
N
g(t) dt = lim g(t) dt.
N→∞
t0
We say that the integral above converges iff the limit exists, and diverges iff the limit does
not exist. In the following Example we compute an improper integral that is very useful to
compute Laplace Transforms.
∞
Example 4.1.1: Compute the improper integral e −at dt, with a ∈ R.
Solution: Following the definition above we need to first compute a definite integral and
then take a limit. So, from the definition,
∞
N
e −at dt.
We first compute the definite integral,
∞
0
0
e −at dt = lim
N→∞
e −at dt = lim
N→∞
Since lim
N→∞ e−aN = 0, for a > 0, we conclude
∞
0
t0
0
0
1
− e
a
−aN
− 1 .
e −at dt = 1
, a > 0. (4.1.1)
a
In the case that a = 0, the improper integral diverges, since we are computing the area of
a rectangle with sides equal to one and infinity. In the case a < 0 holds lim
N→∞ e−aN = ∞.
Therefore, the improper integral converges only for a > 0 and it is given by the expression
in Eq. (4.1.1). ⊳

110 G. NAGY – ODE july 2, 2013
4.1.2. Definition of the Laplace Transform. The Laplace Transform of a function is
defined as an improper integral of the function times an exponential.
Definition 4.1.1. The function ˆ f : Dfˆ → R is the Laplace Transform of the function
f : (0, ∞) → R iff holds
∞
ˆf(s) = e −st f(t) dt (4.1.2)
where D ˆ f ⊂ R is the open set where the integral above converges.
We will often use an alternative notation for ˆ f, the Laplace transform of f, namely,
0
ˆf = L[f].
This notation emphasizes that the Laplace transform defines a map from the set of piecewise
continuous functions into the same set. Maps between function spaces are often called
functionals. Recalling that a scalar-valued function of a single variable is a map denoted
by t ↦→ f(t), the same notation can be used with the Laplace transform, that is, f ↦→ L[f].
In this Section we are interested to study the properties of this map L. In particular we
will show in Theorem 4.1.3 that this map is linear, and in Theorem 4.1.5 that this map is
one-to-one and so invertible on the appropriate sets.
It can be seen from this definition that the Laplace transform of a function f is an improper
integral. We will see in the Examples below that this improper integral in Eq. (4.1.2)
does not converge for every s ∈ (0, ∞). The interval where the Laplace transform of a function
f is well-defined depends on the particular function f. So, we will see below that L[e at ]
with a > 0 is well-defined for s > a, while L[sin(at)] is well-defined for s > 0. We now
compute the Laplace transforms of simple functions.
Example 4.1.2: Compute L[1].
Solution: The function f(t) = 1 is the simplest function we can use to find its Laplace
transform. Following the definition,
∞
L[1] = e −st dt
0
We have computed this improper integral in Example 4.1.1. Just replace a = s in that
example. The result is,
L[1] = 1
s
for s > 0,
and the Laplace Transform is not defined for s 0. ⊳
Example 4.1.3: Compute L[e at ], where a ∈ R.
Solution: Following the definition of a Laplace Transform we obtain,
L[e at ∞
] = e −st e at ∞
dt = e −(s−a)t dt.
0
Here we can again use the result in Example 4.1.1, just replace a in that Example by (s−a).
We obtain,
L[e at 1
] =
for s > a,
(s − a)
and the Laplace Transform is not defined for s a. ⊳
0

Example 4.1.4: Compute L[te at ], where a ∈ R.
G. NAGY – ODE July 2, 2013 111
Solution: In this case the calculation is more complicated than above, since we need to
integrate by parts. We start with the definition of the Laplace Transform,
L[te at ∞
n
] =
te −(s−a)t dt.
We now integrate by parts,
L[te at ] = lim
n→∞
0
e −st te at dt = lim
n→∞
− 1
(s − a) te−(s−a)t
n
0
0
+ 1
s − a
n
0
e −(s−a)t
dt ,
that is,
L[te at
] = lim −
n→∞
1
(s − a) te−(s−a)t
n 1
− e−(s−a)t
0 (s − a) 2 n
0
Notice that the first term on the right-hand side of the last line above vanishes, since
Regarding the other term,
1
lim −
n→∞ (s − a) ne−(s−a)n = 0,
lim
n→∞ −
Therefore, we conclude that
1
(s − a) 2 e−(s−a)n = 0,
L[te at ] =
1
(s − a) te−(s−a)t t=0 = 0.
1
(s − a) 2 e−(s−a)t t=0 =
1
, s > a.
(s − a) 2
Example 4.1.5: Compute L[sin(at)], where a ∈ R.
Solution: In this case we need to compute
L[sin(at)] = lim
n→∞
n
0
e −st sin(at) dt.
1
.
(s − a) 2
The definite integral in the last line above can be computed as follows: Integrating by parts
twice it is not difficult to check that the following formula holds,
n
0
e −st sin(at) dt = − 1
−st
e sin(at)
s
n
−
0
a
s2 −st
e cos(at) n
−
0
a
s2 which implies that
1 + a
s2 n
Hence, it is not difficult to see that
2 2 s + a
∞
which is equivalent to
0
e −st sin(at) dt = − 1
−st
e sin(at)
s
n
−
0
a
s2 s 2
0
L[sin(at)] =
e −st sin(at) dt = a
, s > 0,
s2 a
s2 , s > 0.
+ a2 n
0
e −st sin(at) dt,
−st
e cos(at) n
.
We here present a short list of Laplace transforms. They can be computed in the same
way we computed the examples above.
0
⊳
⊳

112 G. NAGY – ODE july 2, 2013
f(t)
f(t) = 1
ˆ f(s) = L[f(t)]
ˆ f(s) = 1
s
f(t) = e at ˆ f(s) =
1
(s − a)
f(t) = t n ˆ f(s) = n!
s (n+1)
f(t) = sin(at)
f(t) = cos(at)
f(t) = sinh(at)
f(t) = cosh(at)
ˆ f(s) =
ˆ f(s) =
ˆ f(s) =
ˆ f(s) =
f(t) = t n e at ˆ f(s) =
f(t) = e at sin(bt)
f(t) = e at cos(bt)
ˆ f(s) =
ˆ f(s) =
a
s 2 + a 2
s
s 2 + a 2
a
s 2 − a 2
s
s 2 − a 2
n!
(s − a) (n+1)
b
(s − a) 2 + b 2
(s − a)
(s − a) 2 + b 2
s > 0
s > max{a, 0}
s > 0
s > 0
s > 0
s > |a|
s > |a|
Table 2. List of particular Laplace Transforms.
s > max{a, 0}
s > max{a, 0}
s > max{a, 0}
We have seen how to compute the Laplace transform of several functions. We now answer
the following question: How big is the set of functions that have a well-defined Laplace
transform? It turns out that this is a big set of functions, containing all functions that can
be bounded by an exponential. This is the main concept is the following result.
Theorem 4.1.2 (Sufficient conditions). If the function f : R+ → R is piecewise continuous
and there exist positive constants k and a such that
|f(t)| k e at ,
then the Laplace transform of f exists for all s > a.
Proof of Theorem 4.1.2: We show that ˆ f is well-defined, that is, finite, by bounding its
absolute value by a finite number. One can check that the following inequalities hold,
| ˆ
∞
f(s)| = e −st ∞
f(t) dt
e −st ∞
|f(t)| dt e −st ke at dt = k L[e at ].
0
Therefore, we can bound | ˆ f(s)| by the function
0
| ˆ f(s)| k L[e at ] = k
s − a
0
s > a.
Since | ˆ f(s)| is bounded by a finite number for s > a, then ˆ f(s) is well-defined. This bound
establishes the Theorem.

G. NAGY – ODE July 2, 2013 113
4.1.3. Properties of the Laplace Transform. We conclude this Section presenting three
important properties of the Laplace transform, which are summarized in the following three
Theorems. The first one says that the Laplace transform of a linear combination of functions
is the linear combination of their Laplace transforms.
Theorem 4.1.3 (Linear combination). If the Laplace transforms L[f] and L[g] of the
functions f and g are well-defined and a, b are constants, then the following equation holds
L[af + bg] = a L[f] + b L[g].
Proof of Theorem 4.1.3: Since integration is a linear operation, so is the Laplace transform,
as this calculation shows,
∞
L[af + bg] = e
0
−st af(t) + bg(t) dt
∞
= a e −st ∞
f(t) dt + b e −st g(t) dt
0
= a L[f] + b cL[g].
This establishes the Theorem.
Example 4.1.6: Compute L[3t 2 + 5 cos(4t)].
Solution: From the Theorem above we know that
Therefore,
L[3t 2 + 5 cos(4t)] = 3 L[t 2 ] + 5 L[cos(4t)]
2
= 3
s3
4
+ 5
s2 + 42
= 6 20
+
s3 s2 .
+ 42 L[3t 2 + 5 cos(4t)] = 20s3 + 6s2 + 96
s3 (s2 .
+ 16)
The second Theorem relates L[f ′ ] with L[f]. This result is crucial to use the Laplace
transform to solve differential equations.
Theorem 4.1.4 (Derivatives). If the L[f] and L[f ′ ] are well-defined, then holds
L[f ′ ] = s L[f] + f(0).
Furthermore, if L[f ′′ ] is well-defined, then it also holds
L[f ′′ ] = s 2 L[f] − s f(0) − f ′ (0).
Proof of Theorem 4.1.4: The first formula in the Theorem is proved by an integration
by parts, namely,
L[f ′ ] = lim
n→∞
n
e
0
−st f ′ (t) dt
n
= lim e
n→∞
−st n
f(t) − (−s)e
0 0
−st
f(t) dt
= lim e
n→∞
−sn ∞
f(n) − f(0) + s e −st f(t) dt
= −f(0) + s L[f].
0
0
⊳

114 G. NAGY – ODE july 2, 2013
Since L[f] is well-defined, we have used the equation
lim
n→∞ e−snf(n) = 0.
we then conclude that
L[f ′ ] = s L[f] − f(0).
This proves the first part of the Theorem. The furthermore part is obtained using the
relation we have just proved, namely
L[f ′′ ] = s L[f ′ ] − f ′ (0)
= s s L[f] − f(0) − f ′ (0)
= s 2 L[f] − s f(0) − f ′ (0).
This establishes the Theorem.
The third and last Theorem of this Section is also crucial to use this transform to solve
differential equations. It says that the map L acting on the set of piecewise continuous
functions is one-to-one. This property is essential to define the inverse of the Laplace
transform, L −1 . We present this result without the proof.
Theorem 4.1.5 (One-to-one). Given any piecewise continuous functions f and g, it holds
L[f] = L[g] ⇒ f = g.
As we mentioned before, this last result allows to define the inverse Laplace transform:
Given a continuous function with values ˆ f(s), we define L −1 as follows,
L −1 [ ˆ f(s)] = f(t) ⇔ L[f(t)] = ˆ f(s).

4.1.4. Exercises.
4.1.1.- . 4.1.2.- .
G. NAGY – ODE July 2, 2013 115

116 G. NAGY – ODE july 2, 2013
4.2. The initial value problem
4.2.1. Summary of the Laplace Transform method. We use the Laplace Transform
to find solutions to second order differential equations. On the one hand this method works
only with constant coefficients equations, which is a restriction. On the other hand, this
method works with very general source functions, including discontinuous step functions
and Dirac’s delta generalized functions. The main idea of the method can be summarized
in the following scheme:
L
differential eq.
for y(t).
(1)
−→
Algebraic eq.
for L[y(t)].
(2)
−→
Solve the
algebraic eq.
for L[y(t)].
(a) Compute the Laplace transform of the whole differential equation;
(b) Use Theorems 4.1.3 and 4.1.4, that is,
L[a f(t) + b g(t)] = a L[f(t)] + b L[g(t)],
(3)
−→
L y (n) = s n L[y] − s (n−1) y(0) − s (n−2) y ′ (0) − · · · − y (n−1) (0),
Transform back
to obtain y(t).
(Use the table.)
to transform the original differential equation for the unknown function y into an algebraic
equation for the unknown function L[y];
(c) Solve the algebraic equation for L[y];
(d) Transform back to obtain the solution function y.
We provides several examples to help understand how this method works.
Example 4.2.1: Use the Laplace transform to find the solution y to the initial value problem
y ′ + 2y = 0, y(0) = 3.
Solution: We have seen in Section 1.1 that the solution to this problem is y(t) = 3e −2t .
We now recover this result using Laplace Transforms. First, compute the Laplace Transform
of the differential equation,
L[y ′ + 2y] = L[0] = 0.
Theorem 4.1.3 says the Laplace Transform is a linear operation, that is,
L[y ′ ] + 2 L[y] = 0.
Theorem 4.1.4 relates derivatives and multiplications, as follows,
s L[y] − y(0) + 2 L[y] = 0 ⇒ (s + 2)L[y] = y(0).
In the last equation we have been able to transform the original differential equation for y
into an algebraic equation for L[y]. We can solve for the unknown L[y] as follows,
L[y] = y(0)
s + 2
⇒
3
L[y] =
s + 2 ,
where in the last step we introduced the initial condition y(0) = 3. From the list of Laplace
transforms given in Sect. 4.1 we know that
L[e at ] = 1
s − a ⇒
3
s + 2 = 3L[e−2t ] ⇒
3
s + 2 = L[3e−2t ].
So we arrive at L[y] = L[3e −2t ]. The Laplace Transform is invertible in the set of piecewise
continuous functions, then
y(t) = 3e −2t .

G. NAGY – ODE July 2, 2013 117
Example 4.2.2: Use the Laplace transform to find the solution y to the initial value problem
y ′′ − y ′ − 2y = 0, y(0) = 1, y ′ (0) = 0.
Solution: First, compute the Laplace transform of the differential equation,
L[y ′′ − y ′ − 2y] = L[0] = 0.
Theorem 4.1.3 says that the Laplace Transform is a linear operation,
L[y ′′ ] − L[y ′ ] − 2 L[y] = 0.
Then, Theorem 4.1.4 relates derivatives and multiplications,
s 2 L[y] − s y(0) − y ′
(0) − s L[y] − y(0) − 2 L[y] = 0,
which is equivalent to the equation
(s 2 − s − 2) L[y] = (s − 1) y(0) + y ′ (0).
Once again we have transformed the original differential equation for y into an algebraic
equation for L[y]. Introduce the initial condition into the last equation above, that is,
Solve for the unknown L[y] as follows,
(s 2 − s − 2) L[y] = (s − 1).
L[y] =
(s − 1)
(s 2 − s − 2) .
We now use partial fraction methods to find a function whose Laplace transform is the righthand
side of the equation above. First find the roots of the polynomial in the denominator,
s 2 − s − 2 = 0 ⇒ s± = 1
√ s+ = 2,
1 ± 1 + 8 ⇒
2
s− = −1,
that is, the polynomial has two real roots. In this case we factorize the denominator,
L[y] =
(s − 1)
(s − 2)(s + 1) .
The partial fraction decomposition of the right-hand side in the equation above is the following:
Find constants a and b such that
A simple calculation shows
(s − 1) a b
= +
(s − 2)(s + 1) s − 2 s + 1 .
(s − 1) a b a(s + 1) + b(s − 2)
= + =
(s − 2)(s + 1) s − 2 s + 1 (s − 2)(s + 1)
Hence constants a and b must be solutions of the equations
(s − 1) = s(a + b) + (a − 2b) ⇒
a + b = 1,
a − 2b = −1.
The solution is a = 1
3
2
and b = . Hence,
3
L[y] = 1
3
1 2
+
(s − 2) 3
1
(s + 1) .
s(a + b) + (a − 2b)
= .
(s − 2)(s + 1)
⊳

118 G. NAGY – ODE july 2, 2013
From the list of Laplace transforms given in Sect. 4.1 we know that
L[e at ] = 1
s − a ⇒
1
s − 2 = L[e2t ],
1
s + 1 = L[e−t ].
So we arrive at the equation
L[y] = 1
3 L[e2t ] + 2
3 L[e−t
1
2t −t
] = L e + 2e
3
We conclude that
y(t) = 1
2t −t
e + 2e
3
.
Example 4.2.3: Use the Laplace transform to find the solution y to the initial value problem
y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′ (0) = 1.
Solution: First, compute the Laplace transform of the differential equation,
L[y ′′ − 4y ′ + 4y] = L[0] = 0.
Theorem 4.1.3 says that the Laplace Transform is a linear operation,
L[y ′′ ] − 4 L[y ′ ] + 4 L[y] = 0.
Theorem 4.1.4 relates derivatives with multiplication,
s 2 L[y] − s y(0) − y ′
(0) − 4 s L[y] − y(0) + 4 L[y] = 0,
which is equivalent to the equation
(s 2 − 4s + 4) L[y] = (s − 4) y(0) + y ′ (0).
Introduce the initial conditions y(0) = 1 and y ′ (0) = 1 into the equation above,
Solve the algebraic equation for L[y],
(s 2 − 4s + 4) L[y] = s − 3.
L[y] =
(s − 3)
(s 2 − 4s + 4) .
We now want to find a function y whose Laplace transform is the right-hand side in the
equation above. In order to see if partial fraction methods will be needed, we now find the
roots of the polynomial in the denominator,
s 2 − 4s + 4 = 0 ⇒ s± = 1
√
4 ± 16 − 16 ⇒ s+ = s− = 2.
2
that is, the polynomial has a single real root, so L[y] can be written as
L[y] =
(s − 3)
.
(s − 2) 2
This expression is already in the partial fraction decomposition. We now rewrite the righthand
side of the equation above in a way it is simple to use the Laplace Transform table in
Section 4.1,
L[y] =
(s − 2) + 2 − 3
(s − 2) 2
= (s − 2)
−
(s − 2) 2
1
1
⇒ L[y] =
(s − 2) 2 s − 2 −
From the list of Laplace Transforms given in Sect. 4.1 we know that
L[e at ] = 1
s − a ⇒
1
s − 2 = L[e2t ],
1
.
(s − 2) 2
⊳

L[te at ] =
So we arrive at the equation
G. NAGY – ODE July 2, 2013 119
1
(s − a) 2
⇒
1
(s − 2) 2 = L[te2t ].
L[y] = L[e 2t ] − L[te 2t ] = L e 2t − te 2t ⇒ y(t) = e 2t − te 2t .
Example 4.2.4: Use the Laplace transform to find the solution y to the initial value problem
y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′ (0) = 1.
Solution: First, compute the Laplace transform of the differential equation,
L[y ′′ − 4y ′ + 4y] = L[3 sin(2t)].
The right-hand side above can be expressed as follows,
L[3 sin(2t)] = 3 L[sin(2t)] = 3
2
s2 6
=
+ 22 s2 + 4 .
Theorem 4.1.3 says that the Laplace Transform is a linear operation,
L[y ′′ ] − 4 L[y ′ ] + 4 L[y] = 6
s 2 + 4 ,
and Theorem 4.1.4 relates derivatives with multiplications,
s 2 L[y] − s y(0) − y ′
(0) − 4 s L[y] − y(0) + 4 L[y] = 6
s2 + 4 .
Re-order terms,
(s 2 − 4s + 4) L[y] = (s − 4) y(0) + y ′ (0) + 6
s 2 + 4 .
Introduce the initial conditions y(0) = 1 and y ′ (0) = 1,
Solve this algebraic equation for L[y], that is,
L[y] =
(s 2 − 4s + 4) L[y] = s − 3 + 6
s 2 + 4 .
(s − 3)
(s2 − 4s + 4) +
6
(s2 − 4 + 4)(s2 + 4) .
From the Example above we know that s 2 − 4s + 4 = (s − 2) 2 , so we obtain
L[y] = 1
s − 2 −
From the previous example we know that
1
+
(s − 2) 2
6
(s − 2) 2 (s2 . (4.2.1)
+ 4)
L[e 2t − te 2t ] = 1
s − 2 −
1
. (4.2.2)
(s − 2) 2
We know use partial fractions to simplify the third term on the right-hand side of Eq. (4.2.1).
The appropriate partial fraction decomposition for this term is the following: Find constants
a, b, c, d, such that
6
(s − 2) 2 (s2 as + b
=
+ 4) s2 + 4 +
c
(s − 2) +
d
(s − 2) 2
⊳

120 G. NAGY – ODE july 2, 2013
Take common denominator on the right-hand side above, and one obtains the system
a + c = 0,
−4a + b − 2c + d = 0,
4a − 4b + 4c = 0,
4b − 8c + 4d = 6.
The solution for this linear system of equations is the following:
a = 3
,
8
b = 0, c = −3 ,
8
3
d =
4 .
Therefore,
6
(s − 2) 2 (s2 3 s
=
+ 4) 8 s2 3 1 3 1
− +
+ 4 8 (s − 2) 4 (s − 2) 2
We can rewrite this expression above in terms of the Laplace transforms given on the list in
Sect. 4.1, as follows,
6
(s − 2) 2 (s2 3
3
= L[cos(2t)] −
+ 4) 8 8 L[e2t ] + 3
4 L[te2t ],
and using the linearity of the Laplace Transform,
6
(s − 2) 2 (s2
3 3
= L cos(2t) −
+ 4) 8 8 e2t + 3
4 te2t.
(4.2.3)
Finally, introducing Eqs. (4.2.2) and (4.2.3) into Eq. (4.2.1) we obtain
L[y(t)] = L (1 − t) e 2t + 3
8 (−1 + 2t) e2t + 3
8 cos(2t)
.
Since the Laplace Transform is an invertible transformation, we conclude that
y(t) = (1 − t) e 2t + 3
8 (2t − 1) e2t + 3
8 cos(2t).
⊳

4.2.2. Exercises.
4.2.1.- . 4.2.2.- .
G. NAGY – ODE July 2, 2013 121

122 G. NAGY – ODE july 2, 2013
4.3. Discontinuous sources
We have mentioned that the Laplace Transform method is useful to solve differential equations
with discontinuous source functions. In this Section review the simplest discontinuous
function, the step function, and we use it to construct more general piecewise continuous
functions. We then compute the Laplace Transform of these discontinuous functions. We
also establish useful formulas for the Laplace Transform of certain function translations.
These formulas and the Laplace Transform table in Section 4.1 are the key to solve a large
set of differential equations with discontinuous sources.
4.3.1. The Step function. We start with a definition of a step function.
Definition 4.3.1. A function u is called a step function at t = 0 iff holds
0 t < 0,
u(t) =
1 t 0.
The step function u at t = 0 and its right and left translations are plotted in Fig. 14.
u(t)
1
0
t
u(t − c)
1
0 c
t
− c
u(t + c)
Figure 14. The graph of the step function given in Eq. (4.3.1), a right an
a left translation by a constant c > 0, respectively, of this step function.
1
0
(4.3.1)
Recall that given a function with values f(t) and a positive constant c, then f(t − c) and
f(t + c) are the function values of the right translation and the left translation, respectively,
of the original function f. In Fig. 15 we plot the graph of functions f(t) = e at , g(t) = u(t) e at
and their respective right translations by c ∈ R.
a t
a ( t − c )
f ( t ) f ( t ) = e
f ( t ) f ( t ) = e
1
0 t
1
0 c
t
a t
a ( t − c )
f ( t ) f ( t ) = u ( t ) e f ( t ) f ( t ) = u ( t − c ) e
1
0 t
1
0 c
t
Figure 15. An exponential function, its right translation, the function
f(t) = u(t) e at and its right translation.
Right and left translations of step functions are useful to construct bump functions. A
bump function is a function with nonzero values only on a finite interval [a, b]. The simplest
bump function is a box function, b, defined as follows,
⎧
⎪⎨
0 t < a,
b(t) = u(t − a) − u(t − b) ⇔ b(t) = 1
⎪⎩
0
a t < b
t b.
(4.3.2)
In our first Example we plot the graph of a step function using the first expression above,
that is, as a combination of step functions.
t

u(t −a)
1
0 a b
t
G. NAGY – ODE July 2, 2013 123
u(t −b)
1
0 a b
b(t)
1
t 0 a b t
Figure 16. A bump function b constructed with translated step functions.
Example 4.3.1: Graph of the function b(t) = u(t − a) − u(t − b), with 0 < a < b.
Solution: The bump function b can be graphed as follows:
Step and bump functions are useful to construct more general piecewise continuous functions.
Example 4.3.2: Graph of the function f(t) = e at u(t − 1) − u(t − 2) .
Solution:
y
1
a t
f ( t ) = e [ u ( t −1 ) − u ( t −2 ) ]
1 2
e
a t
[ u ( t −1 ) − u ( t −2 ) ]
4.3.2. The Laplace Transform of Step functions. We are interested to find the Laplace
transform of the step functions. The following statement says they are not difficult to
compute.
Theorem 4.3.2. For every real number c and s > 0 holds L[u(t − c)] = e−cs
s .
Proof of Lemma 4.3.2: By the definition of the Laplace Transform,
∞
L[u(t − c)] = e −st ∞
u(t − c) dt = e −st dt,
0
where we used that the step function vanishes for t < c. Now compute the improper integral,
e −Ns − e −cs = e−cs
L[u(t − c)] = lim
N→∞ −1
s
s
⇒ L[u(t − c)] = e−cs
s
This establishes the Theorem.
Example 4.3.3: Compute L[3u(t − 2)].
Solution: The Laplace Transform is a linear operation, so L[3u(t − 2)] = 3 L[u(t − 2)], and
the Theorem above implies that L[3u(t − 2)] = 3e−2s
. ⊳
s
Example 4.3.4: Compute L −1 e3s
.
s
Solution: Since L −1 e3s
= L
s
−1 e−(−3)s
= u(t − (−3)), then, L
s
−1 e3s
= u(t + 3). ⊳
s
c
t
⊳

124 G. NAGY – ODE july 2, 2013
4.3.3. The Laplace Transform of translations. We now introduce two important properties
of the Laplace transform.
Theorem 4.3.3 (Translations). If ˆ f(s) = L[f(t)] exists for s > a 0 and c 0, then
both equations hold,
L[u(t − c)f(t − c)] = e −cs L[f(t)], s > a, (4.3.3)
L[e ct f(t)] = L[f(t)](s − c), s > a + c. (4.3.4)
We can summarize the relations in the Theorem above as follows,
L translation (uf) = (exp) L[f] ,
L (exp) (f) = translation L[f] .
Denoting ˆ f(s) = L[f(t)] we have the following equivalent expression for Eqs. (4.3.3)-(4.3.4),
L[u(t − c)f(t − c)] = e −cs ˆ f(s),
L[e ct f(t)] = ˆ f(s − c).
The inverse form of Eqs. (4.3.3)-(4.3.4) is given by,
L −1 [e −cs ˆ f(s)] = u(t − c)f(t − c), (4.3.5)
L −1 [ ˆ f(s − c)] = e ct f(t). (4.3.6)
Proof of Theorem 4.3.3: The proof is again based in a change of the integration variable.
We start with Eq. (4.3.3), as follows,
∞
L[u(t − c)f(t − c)] = e
0
−st u(t − c)f(t − c) dt
∞
= e
c
−st f(t − c) dt, τ = t − c, dτ = dt
∞
= e
0
−s(τ+c) f(τ) dτ
= e −cs
∞
e −sτ f(τ) dτ
0
= e −cs L[f(t)], s > a.
The proof of Eq. (4.3.4) is even simpler,
L e ct f(t) ∞
= e
0
−st e ct f(t) dt
∞
= e −(s−c)t f(t) dt
0
= L[f(t)](s − c), s − c > a.
This establishes the Theorem.
Example 4.3.5: Compute L u(t − 2) sin(a(t − 2)) .
a
Solution: Both L[sin(at)] =
s2 + a2 and L[u(t − c)f(t − c)] = e−cs L[f(t)] imply
L u(t − 2) sin(a(t − 2)) = e −2s L[sin(at)] = e −2s
We conclude: L u(t − 2) sin(a(t − 2)) =
a
s2 .
+ a2 a e−2s
s2 . ⊳
+ a2

Example 4.3.6: Compute L e 3t sin(at) .
G. NAGY – ODE July 2, 2013 125
Solution: Since L[ectf(t)] = L[f](s − c), then we get
L e 3t sin(at) a
=
(s − 3) 2 , s > 3.
+ a2 Example 4.3.7: Compute both L u(t − 2) cos a(t − 2) and L e3t cos(at) .
Solution: Since L cos(at) s
=
s2 , then
+ a2 L u(t − 2) cos a(t − 2) = e −2s
Example 4.3.8: Find the Laplace transform of the function
0 t < 1,
f(t) =
(t2 − 2t + 2) t 1.
s
(s2 + a2 ) , Le 3t cos(at) (s − 3)
=
(s − 3) 2 .
+ a2 ⊳
⊳
(4.3.7)
Solution: The idea is to re-write function f so we can use the Laplace Transform table in
Section 4.1 to compute its Laplace Transform. Since the function f vanishes for all t < 1,
we use step functions to write f as
f(t) = u(t − 1)(t 2 − 2t + 2).
Now, notice that completing the square we obtain,
t 2 − 2t + 2 = (t 2 − 2t + 1) − 1 + 2 = (t − 1) 2 + 1.
That is, the polynomial is a parabola t 2 translated to the right by 1 and up by one. This is
a discontinuous functions, as it can be seen in Fig. 17.
f ( t )
1
0
1
Figure 17. The graph of the function f given in Eq. (4.3.7).
So the function f can be written as follows,
f(t) = u(t − 1) (t − 1) 2 + u(t − 1).
Since we know that L[t 2 ] = 2
, then Eq. (4.3.3) implies
s3 L[f(t)] = L[u(t − 1) (t − 1) 2 ] + L[u(t − 1)]
−s 2
= e
s
s
3 + e−s 1
t
e−s
⇒ L[f(t)] =
s3 2
2 + s .
⊳

126 G. NAGY – ODE july 2, 2013
Example 4.3.9: Find the function f such that L[f(t)] = e−4s
s 2 + 5 .
Solution: Notice that
L[f(t)] = e −4s
Recall that L[sin(at)] =
1
s 2 + 5
⇒ L[f(t)] = 1
√ 5 e −4s
√ 5
s 2 + √ 5 2 .
a
(s2 + a2 , then Eq. (4.3.3), or its inverse form Eq. (4.3.5) imply
)
L[f(t)] = 1
√ 5 L u(t − 4) sin √ 5 (t − 4) ⇒ f(t) = 1
√ 5 u(t − 4) sin √ 5 (t − 4) .
Example 4.3.10: Find the function f(t) such that L[f(t)] =
Solution: We first rewrite the right-hand side above as follows,
L[f(t)] =
= (s − 2)
(s − 1 − 1 + 1)
(s − 2) 2 + 3
(s − 2) 2 + 3 +
(s − 2)
=
(s − 2) 2 + √ 3 2 1
(s − 2) 2 + 3
+ 1
√ 3
(s − 1)
(s − 2) 2 + 3 .
√ 3
(s − 2) 2 + √ 3 2 .
We now recall Eq. (4.3.4) or its inverse form Eq. (4.3.6), which imply
So, we conclude that
Example 4.3.11: Find L −1 2e−3s s2
.
− 4
Solution: Since L −1 a
s2 − a2
L −1 2e−3s s2
= L
− 4
−1
e −3s 2
s2
− 4
L[f(t)] = L e 2t cos √ 3 t + 1
√ 3 L e 2t sin √ 3 t .
f(t) = e2t
√ √ √
√ 3 cos 3 t + sin 3 t
3
.
Example 4.3.12: Find a function f such that L[f(t)] =
= sinh(at) and L −1 e −cs ˆ f(s) = u(t − c) f(t − c), then
⇒ L −1 2e−3s s2
= u(t − 3) sinh
− 4
2(t − 3) .
e −2s
s 2 + s − 2 .
Solution: Since the right-hand side above does not appear in the Laplace Transform table
in Section 4.1, we need to simplify it in the appropriate way. The plan is to rewrite the
denominator of the rational function 1/(s 2 +s−2), so we can use the partial fraction method
to simplify this rational function. We first find out whether this denominator has real or
complex roots:
s± = 1
√
−1 ± 1 + 8 ⇒
2
s+ = 1,
s− = −2.
⊳
⊳
⊳

We are in the case of real roots, so we can rewrite
G. NAGY – ODE July 2, 2013 127
s 2 + s − 2 = (s − 1) (s + 2).
The partial fraction decomposition in this case is given by
1
(s − 1) (s + 2) =
a
(s − 1) +
b (a + b) s + (2a − b)
=
(s + 2) (s − 1) (s + 2)
The solution is a = 1/3 and b = −1/3, so we arrive to the expression
Recalling that
L[f(t)] = 1
3 e−2s 1 1
−
s − 1 3 e−2s 1
s + 2 .
L[e at ] = 1
s − a ,
and Eq. (4.3.3) we obtain the equation
L[f(t)] = 1
3 L u(t − 2) e (t−2) − 1
3 L u(t − 2) e −2(t−2)
which leads to the conclusion:
f(t) = 1
u(t − 2)
3
e (t−2) − e −2(t−2)
.
⇒
a + b = 0,
2a − b = 1.
4.3.4. Differential equations with discontinuous sources. The last three examples in
this Section show how to use the methods presented above to solve differential equations
with discontinuous source functions.
Example 4.3.13: Use the Laplace Transform to find the solution of the initial value problem
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: We compute the Laplace transform of the whole equation,
L[y ′ ] + 2 L[y] = L[u(t − 4)] = e−4s
s .
From the previous Section we know that
s L[y] − y(0) + 2 L[y] = e −4s
s
⇒ (s + 2) L[y] = y(0) + e−4s
s .
We introduce the initial condition y(0) = 3 into equation above,
L[y] =
3
1
+ e−4s
(s + 2)
s(s + 2) ⇒ L[y] = 3 Le −2t + e −4s 1
s(s + 2) .
We need to invert the Laplace transform on the last term on the right-hand side in equation
above. We use the partial fraction decomposition on the rational function above, as follows
1 a
=
s(s + 2) s +
b a(s + 2) + bs (a + b) s + (2a) a + b = 0,
= = ⇒
(s + 2) s(s + 2) s(s + 2)
2a = 1.
We conclude that a = 1/2 and b = −1/2, so
1 1
=
s(s + 2) 2
1
s −
1
.
(s + 2)
⊳

128 G. NAGY – ODE july 2, 2013
We then obtain
L[y] = 3 L e −2t + 1
−4s 1
e
2 s
= 3 L e −2t + 1
2
Hence, we conclude that
− e−4s
1
(s + 2)
L[u(t − 4)] − L u(t − 4) e −2(t−4)
.
y(t) = 3e −2t + 1
u(t − 4) 1 − e
2 −2(t−4)
.
Example 4.3.14: Use the Laplace transform to find the solution to the initial value problem
y ′′ + y ′ + 5
4 y = b(t), y(0) = 0, y′ (0) = 0, b(t) =
1 0 t < π
0 t π.
Solution: From Fig. 18, the source function g can be written as the following product,
u ( t )
1
0 pi
t
b(t) = u(t) − u(t − π) .
u ( t − pi )
1
0 pi t
b ( t )
1
0 pi t
Figure 18. The graph of the u, its translation and b as given in Eq. (4.3.8).
The last expression for b is particularly useful to find its Laplace transform,
L[b(t)] = L[u(t)] − L[u(t − π)] = 1 1
+ e−πs
s s ⇒ L[b(t)] = (1 − e−πs ) 1
s .
Now Laplace transform the whole equation,
L[y] = L[b].
4
Since the initial condition are y(0) = 0 and y ′ (0) = 0, we obtain
s 2 + s + 5
L[y] =
4
1 − e −πs 1
s ⇒ L[y] = 1 − e −πs 1
s s2 + s + 5
.
4
Introduce the function
1
H(s) =
s s2 + s + 5
4
L[y ′′ ] + L[y ′ ] + 5
⇒ y(t) = L −1 [H(s)] − L −1 [e −πs H(s)].
⊳
(4.3.8)
That is, we only need to find the inverse Laplace Transform of H. We use the partial fraction
method to simplify the expression of H. We first find out whether the denominator has real
or complex roots:
s 2 + s + 5
4 = 0 ⇒ s± = 1
√
−1 ± 1 − 5 ,
2
so the roots are complex-valued. The appropriate partial fraction decomposition is
H(s) =
1
s s2 + s + 5
=
4
a (bs + c)
+
s s2 5 + s + 4

G. NAGY – ODE July 2, 2013 129
Therefore, we get
1 = a s 2 + s + 5
+ s (bs + c) = (a + b) s
4
2 + (a + c) s + 5
4 a.
This equation implies that a, b, and c, satisfy the equations
The solution is, a = 4
5
, b = −4
5
a + b = 0, a + c = 0,
5
a = 1.
4
, c = −4 . Hence, we have found that,
5
1
H(s) =
s2 + s + 5
=
4 s
4
1
5 s −
(s + 1)
s2 5 + s + 4
Complete the square in the denominator,
s 2 + s + 5
4 =
s 2
1
+ 2 s +
2
1
−
4
1 5
+
4 4 =
s + 1
2
Replace this expression in the definition of H, that is,
H(s) = 4
1
5 s −
(s + 1)
1 2
s + 2 + 1
Rewrite the polynomial in the numerator,
(s + 1) =
s + 1
2
1
+ =
2
s + 1
2
+ 1
2 ,
hence we get
H(s) = 4
1
5 s −
s + 1
2
1 2 −
s + 2 + 1 1 1
2 1 2 .
s + 2 + 1
Use the Laplace Transform table to get H(s) equal to
H(s) = 4
L[1] − L
5
e −t/2 cos(t) − 1
2 L[e−t/2
sin(t)] ,
equivalently
Denote:
4
H(s) = L 1 − e
5
−t/2 cos(t) − 1
2 e−t/2
sin(t) .
2
+ 1.
h(t) = 4
1 − e
5
−t/2 cos(t) − 1
2 e−t/2
sin(t) . ⇒ H(s) = L[h(t)].
Recalling L[y(t)] = H(s) + e −πs H(s), we obtain L[y(t)] = L[h(t)] + e −πs L[h(t)], that is,
y(t) = h(t) + u(t − π)h(t − π).
Example 4.3.15: Use the Laplace transform to find the solution to the initial value problem
y ′′ + y ′ + 5
4 y = g(t), y(0) = 0, y′ (0) = 0, g(t) =
sin(t) 0 t < π
0 t π.
Solution: From Fig. 19, the source function g can be written as the following product,
g(t) = u(t) − u(t − π) sin(t),
⊳
(4.3.9)

130 G. NAGY – ODE july 2, 2013
since u(t) − u(t − π) is a box function, taking value one in the interval [0, π] and zero on
the complement. Finally, notice that the equation sin(t) = − sin(t − π) implies that the
function g can be expressed as follows,
g(t) = u(t) sin(t) − u(t − π) sin(t) ⇒ g(t) = u(t) sin(t) + u(t − π) sin(t − π).
The last expression for g is particularly useful to find its Laplace transform,
y
1
sin ( t )
0 pi t
y
1
0
u ( t ) − u ( t − pi )
pi t
y
1
g ( t )
0 pi
t
Figure 19. The graph of the sine function, a square function u(t)−u(t−π)
and the source function g given in Eq. (4.3.9).
L[g(t)] =
1
(s2 1
+ e−πs
+ 1)
(s 2 + 1) .
Having this last transform is then not difficult to solve the differential equation. As usual,
Laplace transform the whole equation,
L[y] = L[g].
4
Since the initial condition are y(0) = 0 and y ′ (0) = 0, we obtain
s 2 + s + 5
L[y] =
4
1 + e −πs 1
Introduce the function
1
H(s) =
s2 + s + 5
4 (s2 + 1)
L[y ′′ ] + L[y ′ ] + 5
(s2 + 1) ⇒ L[y] = 1 + e −πs 1
s 2 + s + 5
⇒ y(t) = L −1 [H(s)] + L −1 [e −πs H(s)].
4
(s2 .
+ 1)
That is, we only need to find the inverse Laplace Transform of H. We use the partial fraction
method to simplify the expression of H. We first find out whether the denominator has real
or complex roots:
s 2 + s + 5
4 = 0 ⇒ s± = 1
√
−1 ± 1 − 5 ,
2
so the roots are complex-valued. The appropriate partial fraction decomposition is
1
(as + b) (cs + d)
H(s) =
s2 5 =
+ s + 4 (s2 + 1) s2 5 +
+ s + (s 4
2 + 1) .
Therefore, we get
1 = (as + b)(s 2
+ 1) + (cs + d) s 2 + s + 5
,
4
equivalently,
1 = (a + c) s 3 + (b + c + d) s 2
+ a + 5
c + d s + b +
4 5
4 d
.
This equation implies that a, b, c, and d, are solutions of
a + c = 0, b + c + d = 0, a + 5
c + d = 0,
4
5
b + d = 1.
4

Here is the solution to this system:
a = 16
17
G. NAGY – ODE July 2, 2013 131
, b = 12
17
4
, c = −16,
d =
17 17 .
We have found that,
H(s) = 4
(4s + 3) (−4s + 1)
17 s2 5 +
+ s + (s 4
2
.
+ 1)
Complete the square in the denominator,
s 2 + s + 5
4 =
s 2
1
+ 2 s +
2
1
−
4
1 5
+
4 4 =
s + 1
2
H(s) = 4
(4s + 3) (−4s + 1)
17
1 2 +
s + 2 + 1 (s2
.
+ 1)
Rewrite the polynomial in the numerator,
(4s + 3) = 4 s + 1 1
− + 3 = 4 s +
2 2
1
+ 1,
2
hence we get
H(s) = 4
1 s + 2
1
s
4
17
1 2 +
1 2 − 4
s + 2 + 1 s + 2 + 1 (s2 + 1) +
1
(s2
.
+ 1)
Use the Laplace Transform table to get H(s) equal to
H(s) = 4
4 L
17
e −t/2 cos(t) + L e −t/2 sin(t)
− 4 L[cos(t)] + L[sin(t)] ,
equivalently
Denote:
h(t) = 4
17
2
+ 1.
4
H(s) = L 4e
17
−t/2 cos(t) + e −t/2
sin(t) − 4 cos(t) + sin(t) .
4e −t/2 cos(t) + e −t/2
sin(t) − 4 cos(t) + sin(t)
⇒ H(s) = L[h(t)].
Recalling L[y(t)] = H(s) + e −πs H(s), we obtain L[y(t)] = L[h(t)] + e −πs L[h(t)], that is,
y(t) = h(t) + u(t − π)h(t − π).
⊳

132 G. NAGY – ODE july 2, 2013
4.3.5. Exercises.
4.3.1.- . 4.3.2.- .

G. NAGY – ODE July 2, 2013 133
4.4. Generalized sources
This Section provides an elementary presentation of a particular generalized function, the
Dirac delta generalized function. We introduce Dirac’s delta using an appropriate sequence
of functions, {δn : R → R}, that is, the sequence domain is the whole set of real numbers.
The limit of the sequence as n → ∞ is not a function R → R though, since the limit diverges
at {0}. Although this limit is a function on the domain R − {0}, it is important to study
the limit of sequences in the same domain where the sequence is defined. And in this same
domain, the limit is not a function. For that reason we call this limit a generalized function,
the Dirac delta generalized function. Moreover, we show that this sequence {δn} has a
well-defined sequence of Laplace Transforms {L[δn]}. What is much more important, the
sequence of Laplace Transforms has a well-defined limit as n → ∞. This is all we need to
solve certain differential equations having the Dirac delta generalized function as a source.
4.4.1. The Dirac delta generalized function. Since this is an elementary presentation
of few generalized functions, we do not mention the theory of distributions, which should
be the most appropriate way to present a complete theory of generalized functions. Since
we restrict ourselves mainly to the Dirac delta generalized function only, we have chosen to
present it as a limit of a function sequence.
A continuous function sequence is a sequence
whose elements are continuous
functions. An example is the continuous
function sequence {un}, for n 1,
is given by
⎧
0, t < 0
⎪⎨
un(t) = nt, 0 t
⎪⎩
1
n
1, t > 1
n .
u
n
1
u u u
3 2 1
0 1/3 1/2 1
Figure 20. The graph of functions
un for n = 1, 2, and 3.
It is not difficult to see that every continuous function can be thought as the limit of a
continuous function sequence. So the set of limits to continuous function sequences contains
the set of all continuous functions. However, the set of limits is much bigger than the set
of continuous functions. For example, it contains discontinuous functions too. The function
sequence {un} is an example of a continuous function sequence whose limit is a discontinuous
function. From the few graphs in Fig. 20 we can infer that limn→∞ un(t) = u(t), the step
function, for t = 0, and the limit does not exist at t = 0. So the limit is a discontinuous
function, although every function in the sequence is continuous. The set of limits include
even stranger objects. May be the most famous of these, which is very important to describe
physical processes, is the Dirac delta generalized function. This object is the limit of certain
continuous function sequences. Since there are many different sequences that have the same
limit and calculations are simple with a discontinuous sequence, we introduce here the Dirac
delta generalized function as the limit of a sequence of bump functions, that is, a sequence
of discontinuous functions.
Definition 4.4.1. Given the sequence for n 1 of bump functions
δn(t) = n u(t) − u t − 1
n
, (4.4.1)
we call the Dirac delta generalized function to the pointwise limit, for every t ∈ R,
lim
n→∞ δn(t) = δ(t).
t

134 G. NAGY – ODE july 2, 2013
It is simple to see that the sequence
of bump functions introduced above
can be re-written as follows,
⎧
0, t < 0
⎪⎨
δn(t) = n, 0 t
⎪⎩
1
n
0, t > 1
n .
We remark that there exist infinitely
many sequences, different for δn
above, that nevertheless define the
same generalized function δ in the
limit n → ∞.
d
n
3
2
1
0
d
3
d
2
1
d
1
Figure 21. The graph of the functions
δn for n = 1, 2, and 3.
We see that the Dirac delta generalized function is a function on the domain R−{0}, indeed
it is the zero function, and it is not defined at t = 0, since the limit diverges at that point.
From the graphs given in Fig. 21 is simple to see that
1
0
t
δn(t) dt = 1 for every n 1.
Therefore, the sequence of integrals is the constant sequence, {1, 1, · · · }, which has a trivial
limit equal to one as n → ∞. This says that the divergence at t = 0 of the sequence
{δn} is of a very particular type. Using this limit procedure, one can generalize several
operations from functions to limits of function sequences. For example, translations, linear
combinations, and multiplications of a function by a generalized function, integration and
Laplace Transforms, as follows:
Definition 4.4.2. We introduce the following operations on the Dirac delta:
f(t) δ(t − c) + g(t) δ(t − c) = lim f(t) δn(t − c) + g(t) δn(t − c) ,
b
a
δ(t − c) dt = lim
n→∞
b
a
n→∞
δn(t − c) dt, L[δ(t − c)] = lim
n→∞ L[δn(t − c)].
Remark: The notation in the definitions above could be misleading. In the left hand
sides above we use the same notation as we use on functions, although Dirac’s delta is not
a function on R. Take the integral, for example. When we integrate a function f, the
integration symbol means “take a limit of Riemann sums”, that is,
b
a
f(t) dt = lim
n
n→∞
i=0
f(xi) ∆x, xi = a + i
b − a
, ∆x =
n n .
However, when f is a generalized function in the sense of a limit of a sequence of functions
{fn}, then by the integration symbol we mean to compute a different limit,
b
a
f(t) dt = lim
n→∞
b
a
fn(t) dt.
We use the same symbol, the integration, to mean two different things, depending whether
we integrate a function or a generalized function. This remark also holds for all the operations
we introduce on generalized functions, specially the Laplace Transform, that will be
often used in the rest of the Section.

G. NAGY – ODE July 2, 2013 135
4.4.2. The Laplace Transform of the Dirac delta. Once we have the definitions of
operations involving the Dirac delta, we can actually compute these limits. The following
statement summarizes few interesting results.
Theorem 4.4.3 (Dirac’s delta). If f : (a, b) → R is continuous and c ∈ (a, b), then
b
a
δ(t − c) dt = 1,
b
a
f(t) δ(t − c) dt = f(c), L[δ(t − c)] = e −cs . (4.4.2)
Proof of Theorem 4.4.3: The first equation in (4.4.2) is simple to see from Fig. 21 and
it is left as an exercise. The second equation can be proven as follows.
b
a
δ(t − c) f(t) dt = lim
n→∞
= lim
n→∞
= lim
n→∞
b
a
b
a
δn(t − c) f(t) dt
n
c+ 1
n
c
u(t − c) − u t − c − 1
n
n f(t) dt,
f(t), dt
where in the last line we used that c ∈ [a, b]. If we denote by F any primitive of f, that is,
F ′ = f, then we can write,
b
a
δ(t − c) f(t) dt = lim
n→∞ nF c + 1
− F (c)
n
= lim
n→∞
1
F c + 1
− F (c)
n
1
n
= F ′ (c) ⇒
b
The third equation in (4.4.2) can be proven as follows.
a
δ(t − c) f(t) dt = f(c).
L[δ(t − c)] = lim
n→∞ L[δn(t − c)]
= lim
n→∞ n
L[u(t − c)] − L u t − c − 1
n
= lim
n→∞ n
1
−cs e e−(c+ n
−
s )s
s
= e −cs s − (1 − e n )
lim
n→∞ s
.
n
The limit on the last line above is a singular limit of the form 0
0 , so we can use the l’Hôpital
rule to compute it, that is,
s − (1 − e n ) −
lim
n→∞ s
= lim
n→∞
n
s
s
e− n
n2
− s
n2 s
= lim e− n = 1.
n→∞
We therefore conclude that
L[δ(t − c)] = e −cs .
This establishes the Theorem.

136 G. NAGY – ODE july 2, 2013
4.4.3. The fundamental solution. In this Subsection we solve differential equations with
the Dirac delta as a source. Since Dirac’s delta is not a function, we need to clarify a bit
what do we mean by a solution to a differential equation with such a source.
Definition 4.4.4. Given constants a1, a0, y0, y1, c ∈ R, consider the sequence of functions
{yn}, with n 1, solutions of the initial value problems
y ′′
n + a1 y ′ n + a0 yn = δn(t − c), yn(0) = y0, y ′ n(0) = y1. (4.4.3)
We define y(t) = L −1
lim
n→∞ L[yn(t)]
to be the solution of the IVP with Dirac’s delta source
y ′′ + a1 y ′ + a0 y = δ(t − c), y(0) = y0, y ′ (0) = y1, (4.4.4)
The definition above seems difficult to work with. However, the following result says that it
is fairly simple to find the function y solution of differential equations with Dirac’s deltas.
Theorem 4.4.5. The function y is solution of the IVP with Dirac’s delta source
y ′′ + a1 y ′ + a0 y = δ(t − c), y(0) = y0, y ′ (0) = y1,
iff its Laplace Transform satisfies the equation
+ a1 s L[y] − y0 − a0 L[y] = e −cs .
s 2 L[y] − sy0 − y 1
This Theorem tells us that to find the solution y to an IVP when the source is a Dirac’s delta
we have to apply the Laplace Transform to the equation and perform the same calculations
as if the Dirac’s delta were a function. So, the final calculation is remarkably simple.
Proof of Theorem 4.4.5: Compute the Laplace Transform on Eq. (4.4.3),
L[y ′′
n] + a 1 L[y ′ n] + a 0 L[yn] = L[δn(t − c)].
Recall the relations between the Laplace Transform and derivatives and use the initial
conditions,
L[y ′′
n] = s 2 L[yn] − sy0 − y1, L[y ′ ] = s L[yn] − y0,
and use these relation in the differential equation,
(s 2 + a1s + a0) L[yn] − sy0 − y1 − a1y0 = L[δn(t − c)],
Since we have shown that lim
n→∞ L[δn(t − c)] = e −cs , then
(s 2 + a1s + a0) lim
n→∞ L[yn] − sy0 − y1 − a1y0 = e −cs .
From the definition of y we see that we can re-write the equation above as
(s 2 + a1s + a0) L[y] − sy0 − y1 − a1y0 = e −cs ,
which is equivalent to
2
s L[y] − sy0 − y1 + a1 s L[y] − y0 − a0 L[y] = e −cs .
This establishes the Theorem.
We now introduce a name for the solution of a differential equation with a Dirac’s delta
generalized function as a source and zero initial conditions. Such solution will play an
important role later on, that is why we give that function a particular name.
Definition 4.4.6. The function yδ is a fundamental solution of the linear operator
L(y) = y ′′ + p y ′ + q y, where p, q are continuous functions, iff the function yδ is solution of
the initial value problem
L(yδ) = δ(t − c), yδ(0) = 0, y ′ δ(0) = 0, c ∈ R.

G. NAGY – ODE July 2, 2013 137
Example 4.4.1: Find the function yδ, the fundamental solution of the initial value problem
y ′′ + 2y ′ + 2y = δ(t), y(0) = 0, y ′ (0) = 0.
Solution: We compute the Laplace transform on both sides of the equation above,
L[y ′′ ] + 2 L[y ′ ] + 2 L[y] = L[δ(t)] = 1 ⇒ (s 2 + 2s + 2) L[y] = 1,
where we have introduced the initial conditions on the last equation above. So we obtain
1
L[y] =
(s2 + 2s + 2) .
The denomination in the equation above has complex valued roots, since
s± = 1
√
−2 ± 4 − 8 ,
2
therefore, we complete squares s2 + 2s + 2 = (s + 1) 2 + 1. We need to solve the equation
L[y] =
1
(s + 1) 2 + 1 = L[e −t sin(t)] ⇒ yδ(t) = e −t sin(t).
Example 4.4.2: Find the solution y to the initial value problem
y ′′ − y = −20 δ(t − 3), y(0) = 1, y ′ (0) = 0.
Solution: The Laplace transform of the differential equation is given by
L[y ′′ ] − L[y] = −20 L[δ(t − 3)] ⇒ (s 2 − 1) L[y] − s = −20 e −3s ,
where in the second equation we have already introduced the initial conditions. We arrive
to the equation
s
L[y] =
(s2 1
− 20 e−3s
− 1) (s2 = L[cosh(t)] − 20 L[u(t − 3) sinh(t − 3)],
− 1)
which leads to the solution
y(t) = cosh(t) − 20 u(t − 3) sinh(t − 3).
Example 4.4.3: Find the solution to the initial value problem
y ′′ + 4y = δ(t − π) − δ(t − 2π), y(0) = 0, y ′ (0) = 0.
Solution: We again Laplace transform both sides of the differential equation,
L[y ′′ ] + 4 L[y] = L[δ(t − π)] − L[δ(t − 2π)] ⇒ (s 2 + 4) L[y] = e −πs − e −2πs ,
where in the second equation above we have introduced the initial conditions. Then,
L[y] = e−πs
(s2 e−2πs
−
+ 4) (s2 + 4)
= e−πs 2
2 (s2 e−2πs 2
−
+ 4) 2 (s2 + 4)
= 1
2 L
u(t − π) sin 2(t − π)
− 1
2 L
u(t − 2π) sin 2(t − 2π)
.
The last equation can be rewritten as follows,
y(t) = 1
2 u(t − π) sin2(t − π) − 1
2 u(t − 2π) sin2(t − 2π) ,
⊳
⊳

138 G. NAGY – ODE july 2, 2013
which leads to the conclusion that
y(t) = 1
u(t − π) − u(t − 2π) sin(2t).
2
Remark: Dirac’s delta generalized function is very useful to describe certain situation in
physical systems. One example is to describe impulsive forces in mechanical systems. An
impulsive force on a particle is a force that transmits a finite momentum to the particle in
an infinitely short time. For example, the momentum transmitted to a pendulum when hit
by a hammer. The force acting on the pendulum due to the hammer can be described with
a Dirac’s delta. Recall Newton’s law of motion,
m v ′ (t) = F (t), with F (t) = F0 δ(t − t0).
Then, the linear momentum transfer from the hammer to the pendulum is,
∆I = lim
∆t→0 mv(t)
t0+∆t
t0+∆t
= lim F (t) dt = F0.
∆t→0
That is, ∆I = F 0.
t0−∆t
t0−∆t
⊳

4.4.4. Exercises.
4.4.1.- . 4.4.2.- .
G. NAGY – ODE July 2, 2013 139

140 G. NAGY – ODE july 2, 2013
4.5. Convolution solutions
This Section describes a way to express a solution to an inhomogeneous linear differential
equation in terms of a convolution between the fundamental solution and the source
function of the differential equation. This particular form to express the solution is useful
in mathematical physics to study different properties of solution. It is also a widely used
in physics and engineering. The plan for this Section is to first present the definition of a
convolution of two functions, and then discuss the main properties of this operation, with
Theorems 4.5.3 and 4.5.5 containing perhaps the most important results of this Section. The
latter result uses convolutions to write solutions to inhomogeneous differential equations.
4.5.1. The convolution of two functions. We start with a definition of the convolution
of two functions. One can say that the convolution is a generalization of the product of two
functions. In a convolution one multiplies the two functions evaluated at different points,
and then integrates the result. Here is a precise definition.
Definition 4.5.1. Assume that f and g are piecewise continuous functions, or one of them
is a Dirac’s delta generalized function. The convolution of f and g is a function denoted
by f ∗ g and given by the following expression,
(f ∗ g)(t) =
t
0
f(τ)g(t − τ) dτ.
In the Example we explicitly compute the integral defining the convolution of two functions.
Example 4.5.1: Find the convolution of the functions f(t) = e −t and g(t) = sin(t).
Solution: The definition of convolution is,
(f ∗ g)(t) =
t
0
e −τ sin(t − τ) dτ.
This integral is not difficult to compute. Integrate by parts twice,
t
e −τ
sin(t − τ) dτ = e −τ t
cos(t − τ) − e −τ t
sin(t − τ) −
0
that is,
2
t
0
e −τ sin(t − τ) dτ =
We then conclude that
0
0
t
0
e −τ sin(t − τ) dτ,
e −τ t
cos(t − τ) − e
0
−τ t
sin(t − τ) = e
0
−t − cos(t) − 0 + sin(t).
(f ∗ g)(t) = 1
−t
e + sin(t) − cos(t) . (4.5.1)
2
⊳
The main properties of the convolution are summarized in the following result:
Lemma 4.5.2 (Properties). For every piecewise continuous functions f, g, and h, hold:
(i) Commutativity: f ∗ g = g ∗ f;
(ii) Associativity: f ∗ (g ∗ h) = (f ∗ g) ∗ h;
(iii) Distributivity: f ∗ (g + h) = f ∗ g + f ∗ h;
(iv) Neutral element: f ∗ 0 = 0;
(v) Identity element: f ∗ δ = f.

G. NAGY – ODE July 2, 2013 141
Proof of Lemma 4.5.2: We only prove properties (i) and (v), the rest are simple to obtain
from the definition of convolution. The first property can be obtained by a change of the
integration variable as follows,
(f ∗ g)(t) =
=
=
t
0
0
t
t
0
f(τ) g(t − τ) dτ, ˆτ = t − τ, dˆτ = −dτ,
f(t − ˆτ) g(ˆτ)(−1) dˆτ
g(ˆτ) f(t − ˆτ) dˆτ ⇒ (f ∗ g)(t) = (g ∗ f)(t).
Finally, property (v) is straightforward form the definition of the delta generalized function,
(f ∗ δ)(t) =
t
0
f(τ) δ(t − τ) dτ = f(t).
This establishes the Lemma.
4.5.2. The Laplace Transform of a convolution. We now show the relation between
the Laplace transform and the convolution operations. This is one of the most important
results we need to solve differential equations. The Laplace transform of a convolution is
the usual product of the individual Laplace transforms.
Theorem 4.5.3 (Laplace Transform). If the functions f and g have well-defined Laplace
Transforms L[f] and L[g], including the case that one of them is a Dirac’s delta, then
L[f ∗ g] = L[f] L[g].
Example 4.5.2: Compute the Laplace transform of the function u(t) =
Solution: The function u above is the convolution of the functions
f(t) = e −t , g(t) = sin(t),
that is, u = f ∗ g. Therefore, Theorem 4.5.3 says that
Since,
L[u] = L[f ∗ g] = L[f] L[g].
L[f] = L[e −t ] = 1
1
, L[g] = L[sin(t)] =
s + 1 s2 + 1 ,
we then conclude that L[u] = L[f ∗ g] is given by
L[f ∗ g] =
1
(s + 1)(s 2 + 1) .
t
0
e −τ sin(t−τ) dτ.
Proof of Theorem 4.5.3: The key step is to interchange two integrals. We start we the
product of the Laplace Transforms,
L[f] L[g] =
∞
e
0
−st
f(t) dt
∞
e
0
−s˜t
g(˜t) d˜t
∞
= e
0
−s˜t
g(˜t)
∞
e
0
−st
f(t) dt d˜t
∞
= g(˜t)
∞
e −s(t+˜t)
f(t) dt d˜t,
0
0
⊳

142 G. NAGY – ODE july 2, 2013
where we only introduced the integral in t as a constant inside the integral in ˜t. Introduce
the Change of variables in the inside integral τ = t + ˜t, hence dτ = dt. Then, we get
∞
L[f] L[g] = g(˜t)
0
∞
e
˜t
−sτ
f(τ − ˜t) dτ d˜t (4.5.2)
∞ ∞
= e −sτ g(˜t) f(τ − ˜t) dτ d˜t. (4.5.3)
Here is the key step, we must switch the order of
integration. From Fig. 22 we see that changing the
order of integration gives the following expression,
∞ τ
L[f] L[g] = e −sτ g(˜t) f(τ − ˜t) d˜t dτ.
0
0
Then, is straightforward to check that
∞
L[f] L[g] = e
0
−sτ τ
g(˜t) f(τ − ˜t) d˜t dτ
0
∞
= e −sτ (g ∗ f)(τ) dt
0
= L[g ∗ f] ⇒ L[f] L[g] = L[f ∗ g].
This establishes the Theorem.
Example 4.5.3: Use Laplace transforms to compute u(t) =
0
˜t
t
0
t = tau
Figure 22. Domain of
integration in (4.5.3).
t
0
e −τ sin(t − τ) dτ.
Solution: We know that u = f ∗ g, with f(t) = e−t and g(t) = sin(t), and we have seen in
Example 4.5.2 that
1
L[u] = L[f ∗ g] =
(s + 1)(s2 + 1) .
A partial fraction decomposition of the right-hand side above implies that
This says that
We then conclude that
L[u] = 1
1 (1 − s)
+
2 (s + 1) (s2
+ 1)
= 1
1
2 (s + 1) +
1
(s2 + 1) −
s
(s2
+ 1)
= 1
L[e
2
−t
] + L[sin(t)] − L[cos(t)] .
u(t) = 1
−t
e + sin(t) − cos(t) .
2
(f ∗ g)(t) = 1
−t
e + sin(t) − cos(t) ,
2
which agrees with Eq. (4.5.1) in the first example above. ⊳
tau

G. NAGY – ODE July 2, 2013 143
4.5.3. Impulse response solutions. We now introduce the notion of the impulse response
solution to a differential equation. This is the solution of the the initial value problem for
the differential equation with zero initial conditions and a Dirac’s delta as a source function.
Definition 4.5.4. The impulse response solution is the function yδ solution of the IVP
y ′′
δ + a1 y ′ δ + a0 yδ = δ(t − c), yδ(0) = 0, y ′ δ(0) = 0, c ∈ R.
Suppose the differential equation describes the motion of a particle and the the source is the
force applied to the particle. The impulse response solution describes the movement of the
particle that starts at the initial position with zero velocity and moves because an impulsive
force kicked the particle in an infinitesimal short time.
Example 4.5.4: Find the impulse response solution of the initial value problem
y ′′
δ + 2 y ′ δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′ δ(0) = 0, .
Solution: We have to use the Laplace Transform to solve this problem because the source
is a Dirac’s delta generalized function. So, compute the Laplace Transform of the differential
equation,
L[y ′′
δ ] + 2 L[y ′ δ] + 2 L[yδ] = L[δ(t − c)].
Since the initial conditions are all zero, we get
Find the roots of the denominator,
(s 2 + 2s + 2) L[yδ] = e −cs ⇒ L[yδ] =
e −cs
(s 2 + 2s + 2) .
s 2 + 2s + 2 = 0 ⇒ s± = 1 √
−2 ± 4 − 8
2
The denominator has complex roots. Then, it is convenient to complete the square in the
denominator,
s 2
+ 2s + 2 = s 2
2
+ 2 s + 1 − 1 + 2 = (s + 1)
2
2 + 1.
Therefore, we obtain the expression,
L[yδ] =
e −cs
(s + 1) 2 + 1 .
Recall that L[sin(t)] = 1
s 2 + 1 , and L[f](s − c) = L[ect f(t)]. Then,
1
(s + 1) 2 + 1 = L[e−t sin(t)] ⇒ L[yδ] = e −cs L[e −t sin(t)].
Since e −cs L[f](s) = L[u(t − c) f(t − c)], we conclude that
yδ(t) = u(t − c) e −(t−c) sin(t − c).
⊳

144 G. NAGY – ODE july 2, 2013
4.5.4. The solution decomposition Theorem. One could say Theorem 4.5.5 is the main
result of this Section. It presents a decomposition of the solution to a general second order,
linear, constant coefficients initial value problem. The statement says that the solution to
such problem can always be split in two terms, the first containing information only about
the initial data, while the second contains information only about the source function.
Theorem 4.5.5 (Solution decomposition). Given constants a 0, a 1, y 0, y 1 and a piecewise
continuous function g, the solution y to the initial value problem
can be decomposed as
y ′′ + a 1 y ′ + a 0 y = g(t), y(0) = y 0, y ′ (0) = y 1, (4.5.4)
y(t) = yh(t) + (yδ ∗ g)(t), (4.5.5)
where yh is the solution of the homogeneous initial value problem
y ′′
h + a 1 y ′ h + a 0 yh = 0, yh(0) = y 0, y ′ h(0) = y 1, (4.5.6)
and yδ is the impulse response solution, that is,
y ′′
δ + a1 y ′ δ + a0 yδ = δ(t), yδ(0) = 0, y ′ δ(0) = 0.
The solution decomposition in Eq. (4.5.5) can be expressed in the equivalent way
y(t) = yh(t) +
t
0
yδ(τ)g(t − τ) dτ.
Proof of Theorem4.5.5: Compute the Laplace Transform of the differential equation,
L[y ′′ ] + a1 L[y ′ ] + a0 L[y] = L[g(t)].
Recalling the relations between Laplace Transforms and derivatives,
L[y ′′ ] = s 2 L[y] − sy0 − y1, L[y ′ ] = s L[y] − y0.
we re-write the differential equation for y as an algebraic equation for cL[y],
(s 2 + a1s + a0) L[y] − sy0 − y1 − a1y0 = L[g(t)].
As usual, it is simple to solve the algebraic equation for cL[y],
L[y] = (s + a1)y0 + y1
(s2 + a1s + a0) +
1
(s2 + a1s + a0) L[g(t)].
Now, the function yh is the solution of Eq. (4.5.6), that is,
L[yh] = (s + a 1)y 0 + y 1
(s 2 + a1s + a0) .
And by the definition of the impulse response solution yδ we have that
1
L[yδ] =
(s2 + a1s + a0) .
These last three equation imply,
L[y] = L[yh] + L[yδ] L[g(t)].
This is the Laplace Transform version of Eq. (4.5.5). Inverting the Laplace Transform above,
y(t) = yh(t) + L −1 L[yδ] L[g(t)] .
Using the result in Theorem 4.5.3 in the last term above we conclude that
y(t) = yh(t) + (yδ ∗ g)(t).

G. NAGY – ODE July 2, 2013 145
Example 4.5.5: Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′ (0) = −1.
Solution: Compute the Laplace Transform of the differential equation above,
L[y ′′ ] + 2 L[y ′ ] + 2 L[y] = L[sin(at)],
and recall the relations between the Laplace Transform and derivatives,
L[y ′′ ] = s 2 L[y] − sy(0) − y ′ (0), L[y ′ ] = s L[y] − y(0).
Introduce the initial conditions in the equation above,
L[y ′′ ] = s 2 L[y] − s (1) − (−1), L[y ′ ] = s L[y] − 1,
and these two equation into the differential equation,
Re-order terms to get
(s 2 + 2s + 2) L[y] − s + 1 − 2 = L[sin(at)].
L[y] =
(s + 1)
(s2 + 2s + 2) +
1
(s2 + 2s + 2) L[sin(at)].
Now, the function yh is the solution of the homogeneous initial value problem with the same
initial conditions as y, that is,
L[yh] =
(s + 1)
(s2 (s + 1)
=
+ 2s + 2) (s + 1) 2 + 1 = L[e−t cos(t)].
Now, the function yδ is the impulse response solution for the differential equation in this
Example, that is,
1
cL[yδ] =
(s2 + 2s + 2) =
1
(s + 1) 2 + 1 = L[e−t sin(t)].
Put all this information together and denote g(t) = sin(at) to get
More explicitly, we get
L[y] = L[yh] + L[yδ] L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),
y(t) = e −t cos(t) +
t
0
e −τ sin(τ) sin[a(t − τ)] dτ.
⊳

146 G. NAGY – ODE july 2, 2013
4.5.5. Exercises.
4.5.1.- . 4.5.2.- .

G. NAGY – ODE July 2, 2013 147
Chapter 5. Systems of differential equations
The main interest of this Chapter resides in its last Section, where we study in some detail
the solutions of 2 × 2 linear first order differential equations with constant coefficients. A
system of differential equations is harder to solve than a single differential equation because
the equations are coupled. The derivative of the i-th unknown depends on all the other
unknowns, not just the i-th one. In the last Section of this Chapter we find solutions to
linear differential systems in the case that the constant coefficients matrix is diagonalizable.
The latter means that when the system is written in terms of new unknowns based on
the coefficient matrix eigenvectors, the system decouples. One then solves the individual
decoupled equations, and then transforms back the result to the original unknowns. In order
to present this method we first need to review well-known material form linear algebra. This
review will focus on on matrix algebra, determinants, linearly dependent and independent
vector sets, and the calculation of eigenvalues and eigenvectors of a square matrix.
5.1. Introduction
5.1.1. Systems of differential equations. A system of differential equations is a set
of more than one differential equation on more than one unknown function. Systems of
differential equations often appear in physical descriptions of nature. We start with may be
the most common example, Newton’s law of motion for point particles.
Example 5.1.1: Newton’s second law of motion for a point particle of mass m moving
in space is a system of differential equations. The unknown functions are the particle
coordinates in space, x 1, x 2, and x 3. The source functions are the three components of the
force, F 1, F 2, and F 3, which depend on time and space. Newton’s second law of motion is
a system of three differential equations given by
m d2x1 dt2 = F
1 t, x1(t), x2(t), x3(t) ,
m d2x2
= F2 t, x1(t), x2(t), x3(t)
dt2 ,
m d2x3
= F3 t, x1(t), x2(t), x3(t)
dt2 .
This is an example of a system of three differential equations in the three unknown functions
x1, x2, and x3. The equations are of second order, since they involve second derivatives of
the unknown functions. We will see in later Sections that it is convenient to group both the
unknown functions and the source functions in vector valued functions,
⎡ ⎤
⎡ ⎤
x1(t) F1(t, x)
x(t) = ⎣x2(t)
⎦ , F(t) = ⎣F2(t,
x) ⎦ .
x3(t)
F3(t, x)
We will then use vector notation to express systems of differential equations. In the case of
Newton’s law of motion we get
m d2 x
dt 2 (t) = F t, x(t) .
Newton’s law of motion mentioned above is a particular case of what is called a second
order differential system. This is our first example of a system of differential equations
because Newton’s law are so widely used in physics. However, in these notes we start
introducing a first order differential system.
⊳

148 G. NAGY – ODE july 2, 2013
Definition 5.1.1. An n × n first order system of differential equations is the following:
Given functions Fi : [t1, t2] × Rn → R, where i = 1, · · · , n, find n functions
xj : [t1, t2] → R solutions of the n linear differential equations
x ′
1(t) = F1 t, x1(t), · · · , xn(t) , (5.1.1)
.
x ′
n(t) = Fn t, x1(t), · · · , xn(t) . (5.1.2)
Example 5.1.2: n = 1: This is a single differential equation: Find x1(t) solution of
x ′ (x1) 1 = −
3
2x1 + t(x1) 2
Actually, this equation can be transformed into an exact equation when multiplied by an
appropriate integrating factor. We have seen how to find solutions solutions to equations of
this type in Section 1.4. ⊳
Example 5.1.3: n = 2: This is a 2 × 2 differential system: Find x1(t) and x2(t) solutions of
x ′ 1 = e t (x1) 2 + t 2 (x2) 2 ,
x ′ 2 = e x1x2 . ⊳
An important particular case of an n × n differential system is when the differential
equations are linear in the unknown functions.
Definition 5.1.2. An n × n first order system of linear differential equations is
the following: Given functions aij, gi : [t1, t2] → R, where i, j = 1, · · · , n, find n functions
xj : [t1, t2] → R solutions of the n linear differential equations
x ′ 1 = a11(t) x1 + · · · + a1n(t) xn + g1(t),
.
x ′ n = an1(t) x 1 + · · · + ann(t) xn + gn(t).
The system is called homogeneous iff the source functions satisfy that g 1 = · · · = gn = 0.
The system is called of constant coefficients iff all the functions aij are constants.
We see that the linear differential system is a particular case of a differential system where
functions Fi, for i = 1, · · · , n have the form
Fi(t, x1, · · · , xn) = ai1(t) x1 + · · · + ain(t) xn + gi(t),
that is, the Fi are linear functions of the xi.
Example 5.1.4: n = 1: This is a single differential equation: Find x1(t) solution of
x ′ 1 = a11(t) x 1 + g 1(t).
This is a linear first order equation, and solutions can be found with the integrating factor
method described in Section 1.2. ⊳
Example 5.1.5: n = 2: This is a 2 × 2 linear system: Find x 1(t) and x 2(t) solutions of
x ′ 1 = a11(t) x1 + a12(t) x2 + g1(t),
x ′ 2 = a21(t) x1 + a22(t) x2 + g2(t). ⊳

G. NAGY – ODE July 2, 2013 149
Example 5.1.6: n = 2: A 2 × 2 homogeneous linear system: Find x 1(t) and x 2(t) such that
x ′ 1 = a11(t) x 1 + a12(t) x 2
x ′ 2 = a21(t) x1 + a22(t) x2. ⊳
A solution of an n × n differential system is a set of n functions {x 1, · · · , xn} that satisfy
every equation in the differential system. Usually it is more complicated to solve a system
of differential equations than a single differential equation. The reason is that in a system
the equations are usually coupled. For example, consider the 2 × 2 system in Example 5.1.6
above. One must know the function x2 in order to integrate the first equation to obtain the
function x1. Similarly, one has to know function x1 to integrate the second equation to get
function x2. This is what we mean by a coupled system. One cannot integrate one equation
at a time; one must integrate the whole system together. We now show how this can be
done in a simple example. Although the example is simple, the idea to find solutions is
powerful. Generalizations of this idea is what one does to solve much more general systems
of differential equations.
Example 5.1.7: Find functions x1, x2 solutions of the first order, 2×2, constant coefficients,
homogeneous differential system
x ′ 1 = x1 − x2,
x ′ 2 = −x1 + x2.
Solution: The main idea to solve this system comes from the following observation. If we
add up the two equations equations, and if we subtract the second equation from the first,
we obtain, respectively,
(x1 + x2) ′ = 0, (x1 − x2) ′ = 2(x1 − x2).
To understand the consequences of what we have done, let us introduce the new unknowns
v = x1 + x2, and w = x1 − x2, and re-write the equations above with these new unknowns,
v ′ = 0, w ′ = 2w.
We have decoupled the original system. The equations for x1 and x2 are coupled, but we
have found a linear combination of the equations such that the equations for v and w are
not coupled. We now solve each equation independently of the other.
v ′ = 0 ⇒ v = c 1,
w ′ = 2w ⇒ w = c2e 2t ,
with c1, c2 ∈ R. Having obtained the solutions for the decoupled system, we now transform
back the solutions to the original unknown functions. From the definitions of v and w we
see that
x1 = 1
(v − w).
2 (v + w), x2 = 1
2
We conclude that for all c1, c2 ∈ R the functions x1, x2 below are solutions of the 2 × 2
differential system in the example, namely,
x 1(t) = 1
2 (c 1 + c 2e 2t ), x 2(t) = 1
2 (c 1 − c 2e 2t ).
This example contains the main idea to solve systems of linear differential equations. If
the differential equations in a linear system are coupled, then find appropriate linear combinations
of the equations and the unknowns such that the new system for new unknowns
⊳

150 G. NAGY – ODE july 2, 2013
is not coupled. Solve each equation in the uncoupled system using the methods from Section
1.2. Once the solution for the new unknowns are obtained, transform them back to the
original unknowns. The resulting functions will be solution to the original linear differential
system.
We key part in this method is to find the appropriate linear combinations of the equations
and unknowns that decouples the system. In the example above that transformation was
simple to find. This is not the case in a general linear system, and this is the reason we will
need to review concepts from Linear Algebra. We will find out that the transformation that
decouples a linear differential system is deeply related with the eigenvalues and eigenvectors
of the system coefficient matrix.
5.1.2. Reduction to first order systems. A second order linear differential equation for
a single unknown function can be transformed into a 2 × 2 linear system for the unknown
and its first derivative. More precisely, we have the following statement.
Theorem 5.1.3 (First order reduction). Every solution y to the second order equation
y ′′ + p(t) y ′ + q(t) y = g(t), (5.1.3)
defines a solution x1 = y and x2 = y ′ of the 2 × 2 first order linear differential system
x ′ 1 = x 2, (5.1.4)
x ′ 2 = −q(t) x1 − p(t) x2 + g(t). (5.1.5)
Conversely, every solution x 1, x 2 of the 2×2 first order linear system in Eqs. (5.1.4)-(5.1.5)
defines a solution y = x 1 of the second order differential equation in (5.1.3).
Proof of Theorem 5.1.3:
(⇒) Given a solution y of Eq. (5.1.3), introduce the functions x1 = y and x2 = y ′ . Therefore
Eq. (5.1.4) holds, due to the relation
x ′ 1 = y ′ = x 2,
Also Eq. (5.1.5) holds, because of the equation
x ′ 2 = y ′′ = −q(t) y − p(t) y ′ + g(t) ⇒ x ′′
2 = −q(t) x1 − p(t) x2 + g(t).
(⇐) Differentiate Eq. (5.1.4) and introduce the result into Eq. (5.1.5), that is,
Denoting y = x1, we obtain,
x ′′
1 = x ′ 2 ⇒ x ′′
1 = −q(t) x1 − p(t) x ′ 1 + g(t).
y ′′ + p(t) y ′ + q(t) y = g(t).
This establishes the Theorem.
Example 5.1.8: Express as a first order system the second order equation
Solution: Introduce the new unknowns
y ′′ + 2y ′ + 2y = sin(at).
x1 = y, x2 = y ′ ⇒ x ′ 1 = x2.
Then, the differential equation can be written as
We conclude that
x ′ 2 + 2x2 + 2x1 = sin(at).
x ′ 1 = x2, x ′ 2 = −2x1 − 2x2 + sin(at).
⊳

G. NAGY – ODE July 2, 2013 151
Example 5.1.9: Express as a single second order equation the 2 × 2 system and solve it,
x ′ 1 = −x1 + 3x2,
x ′ 2 = x1 − x2.
Solution: Compute x1 from the second equation: x1 = x ′ 2 + x2. Introduce this expression
into the first equation,
(x ′ 2 + x2) ′ = −(x ′ 2 + x2) + 3x2 ⇒ x ′′
2 + x ′ 2 = −x ′ 2 − x2 + 3x2,
so we obtain the second order equation
x ′′
2 + 2x ′ 2 − 2x2 = 0.
We solve this equation with the methods studied in Chapter 2, that is, we look for solutions
of the form x2(t) = e rt , with r solution of the characteristic equation
r 2 + 2r − 2 = 0 ⇒ r± = 1 √
−2 ± 4 + 8 ⇒ r± = −1 ±
2
√ 3.
Therefore, the general solution to the second order equation above is x2 = c1 e r+ t + c2 e r− t ,
with c1, c2 ∈ R. Since x1 = x ′ 2 + x2, we obtain
x 1 = c 1r+ e r+ t + c 2r− e r− t + c 1 e r+ t + c 2 e r− t ,
Re-ordering terms we get the expression for the function x1, that is,
x1 = c1(1 + r+) e r+ t + c2(1 + r−) e r− t .
5.1.3. The Picard-Lindelöf Theorem. The Picard-Lindelöf Theorem says that every
initial value problem for a certain class of differential systems always has a solution and
the solution is unique. We start introducing the notion of an initial value problem for
differential systems. This is a straightforward generalization of an initial value problem for
a single equation.
Definition 5.1.4. An initial value problem for the n × n differential system given in
Definition 5.1.1 is the following: Given the set of n + 1 real numbers {x 01, · · · , x 0 n} and t 0,
find a set of n functions {x1, · · · , xn} solutions of the Eqs. (5.1.1)-(5.1.2) together with the
initial conditions
x 1(t 0) = x 01, · · · xn(t 0) = x 0 n.
Since an n × n differential system has n unknown functions, we define the initial value
problem imposing n additional conditions, one for each unknown function.
We now present a version of the Picard-Lindelöf Theorem, which states the following:
If the functions Fi defining the differential system are differentiable both in t and in the
arguments where we place the unknown functions, then the initial value problem in Definition
5.1.4 has a unique solution for every real numbers {x01, · · · , x0 n} and t0. This is not the
strongest version of the Theorem but it is simpler to describe in these notes. The strongest
version requires that Fi be only continuous in t and a condition between continuity and
differentiability, called Lipschitz continuous, in the arguments where we place the unknown
functions. Charles Émile Picard presented in 1890 a successive approximation method to
find solutions to differential equations and Ernst Lindelöf applied this method in 1894 to
find solutions to Eqs. (5.1.1)-(5.1.2).
⊳

152 G. NAGY – ODE july 2, 2013
Theorem 5.1.5 (Picard-Lindelöf). If D ⊂ R×R n is a non-empty open set, and the functions
Fi : D → R are differentiable, i = 1, · · · , n, then for every point (t0, x01, · · · , x0 n) ∈ D
there exists an interval (t1, t2) ⊂ R containing t0 such that there exists a unique solution set
{x1, · · · , xn}, with xi : (t1, t2) → R, of the initial value problem
x ′
1(t) = F1 t, x1(t), · · · , xn(t) , · · · x ′
n(t) = Fn t, x1(t), · · · , xn(t) ,
x 1(t 0) = x 01, · · · xn(t 0) = x 0 n.
Although the proof of this result involves techniques beyond the scope of these notes, the
main ideas of the proof are not difficult to understand, specially using vector notation. That
is why we show these ideas after we finished our review of Linear Algebra.
The Picard-Lindelöf Theorem applies to linear differential systems introduced in Definition
5.1.2, since these systems are a particular case of the differential systems in Definition
5.1.1. Nevertheless, in Section 5.6 we present a simpler proof of Picard-Lindelöf
Theorem for linear, constant coefficients, differential systems in the case that the equation
coefficients satisfy an extra condition, called diagonalizability. This is a rather particular
type of systems but they are fairly common in applications. This simple proof follows the
main idea we used to solve Example 5.1.7. To understand the diagonalizability condition is
why we need to review several notions from Linear Algebra. We dedicate few Sections to
such review.

5.1.4. Exercises.
5.1.1.- . 5.1.2.- .
G. NAGY – ODE July 2, 2013 153

154 G. NAGY – ODE july 2, 2013
5.2. Systems of algebraic equations
We said in the previous Section that one way to solve a linear differential system is to
transform the original system and unknowns into a decoupled system, solve the decoupled
system for the new unknowns, and then transform back the solution to the original unknowns.
These transformations, to and from, the decoupled system are the key steps to
solve a linear differential system. One way to understand these transformations is using the
ideas and notation from Linear Algebra. This Section is a review of concepts from Linear
Algebra needed to introduce the transformations mentioned above. We start introducing
an algebraic linear system, we then introduce matrices and matrix operations, column vectors
and matrix vector products. The transformations on a system of differential equations
mentioned above will be introduced in a later Section, when we study the eigenvalues and
eigenvectors of a square matrix.
5.2.1. Linear algebraic systems. One could say that the set of results we call Linear
Algebra originated with the study of linear systems of algebraic equations. Our review of
elementary concepts from Linear Algebra starts with a study of these type of equations.
Definition 5.2.1. An n × n system of linear algebraic equations is the following:
Given constants aij and bi, where i, j = 1 · · · , n 1, find the constants xj solutions of
a11x1 + · · · + a1nxn = b1, (5.2.1)
.
an1x1 + · · · + annxn = bn. (5.2.2)
The system is called homogeneous iff all sources vanish, that is, b1 = · · · = bn = 0.
Example 5.2.1:
(a) A 2 × 2 linear system on the unknowns x 1 and x 2 is the following:
2x1 − x2 = 0,
−x1 + 2x2 = 3.
(b) A 3 × 3 linear system on the unknowns x1, x2 and x3 is the following:
x 1 + 2x 2 + x 3 = 1,
−3x 1 + x 2 + 3x 3 = 24,
x 2 − 4x 3 = −1. ⊳
One way to find a solution to an n × n linear system is by substitution. Compute x1
from the first equation and introduce it into all the other equations. Then compute x2 from
this new second equation and introduce it into all the remaining equations. Repeat this
procedure till the last equation, where one finally obtains xn. Then substitute back and
find all the xi, for i = 1, · · · , n − 1. A computational more efficient way to find a solution
is to perform Gauss elimination operations on the augmented matrix of the system. Since
matrix notation will simplify calculations, it is convenient we spend some time on this. We
start with the basic definitions.
Definition 5.2.2. An m × n matrix, A, is an array of numbers
⎡
⎤
a11 · · · a1n
⎢
⎥ m rows,
A = ⎣ . . ⎦ ,
aij ∈ C,
n columns,
am1 · · · amn
where i = 1, · · · , m, and j = 1, · · · , n. An n × n matrix is called a square matrix.

G. NAGY – ODE July 2, 2013 155
Example 5.2.2:
(a) Examples of 2 × 2, 2 × 3, 3 × 2 real-valued matrices, and a 2 × 2 complex-valued matrix:
⎡ ⎤
1 2
1 2
1 2 3
A = , B = , C = ⎣3
4⎦
1 + i 2 − i
, D =
.
3 4
4 5 6
3 4i
5 6
(b) The coefficients of the algebraic linear systems in Example 5.2.1 can be grouped in
matrices, as follows,
2x1 − x2 = 0,
2
⇒ A =
−x1 + 2x2 = 3,
−1
−1
.
2
⎫
x1 + 2x2 + x3 = 1,
⎡
⎪⎬
1
−3x1 + x2 + 3x3 = 24, ⇒ A = ⎣−3
⎪⎭
x2 − 4x3 = −1.
0
2
1
1
⎤
1
3 ⎦ .
−4
The particular case of an m × 1 matrix is called an m-vector.
Definition 5.2.3. An m-vector, v, is the array of numbers v =
components vi ∈ C, with i = 1, · · · , m.
⎡
⎢
⎣
v1
.
vm
⎤
⊳
⎥
⎦, where the vector
Example 5.2.3: The unknowns of the algebraic linear systems in Example 5.2.1 can be
grouped in vectors, as follows,
⎫
x1 + 2x2 + x3 = 1,
⎡
⎪⎬
2x1 − x2 = 0,
x1
⇒ x = . −3x1 + x2 + 3x3 = 24, ⇒ x = ⎣
−x1 + 2x2 = 3,
x2
⎪⎭
x2 − 4x3 = −1.
x1
⎤
x2⎦
.
x3
Definition 5.2.4. The matrix-vector product of an n × n matrix A and an n-vector x
is an n-vector given by
⎡
⎤ ⎡ ⎤ ⎡
⎤
Ax =
⎢
⎣
a11 · · · a1n
.
an1 · · · ann
.
⎥ ⎢
⎦ ⎣
x1
.
xn
⎥ ⎢
⎦ = ⎣
a11x1 + · · · + a1nxn
.
an1x1 + · · · + a1nxn
The matrix-vector product of an n × n matrix with an n-vector is another n-vector. This
product is useful to express linear systems of algebraic equations in terms of matrices and
vectors.
Example 5.2.4: Find the matrix-vector products for the matrices A and vectors x in Examples
5.2.2(b) and Example 5.2.3, respectively.
Solution: In the 2 × 2 case we get
2 −1 x1 2x1 − x2
Ax =
=
.
−1 2 x2 −x1 + 2x2
In the 3 × 3 case we get,
⎡
1
Ax = ⎣−3 2
1
⎤ ⎡
1
3 ⎦
0 1 −4
⎣ x1
x2
x3 ⎤ ⎡
⎦ = ⎣ x1
⎤
+ 2x2 + x3
−3x1 + x2 + 3x3⎦
.
x 2 − 4x 3
⎥
⎦
⊳
⊳

156 G. NAGY – ODE july 2, 2013
Example 5.2.5: Use the matrix-vector product to express the algebraic linear system below,
2x1 − x2 = 0,
−x1 + 2x2 = 3.
Solution: Introduce the coefficient matrix A, the unknown vector x, and the source vector
b as follows,
2
A =
−1
−1
,
2
x1
x = ,
0
b = .
3
Since the matrix-vector product Ax is given by
2 −1 x1 2x1 − x2
Ax =
=
,
−1 2 x2 −x1 + 2x2
then we conclude that
2x1 − x2 = 0,
−x1 + 2x2 = 3,
⇔
x2
2x1 − x2
=
−x1 + 2x2
0
3
⇔ Ax = b.
It is simple to see that the result found in the Example above can be generalized to every
n × n algebraic linear system.
Theorem 5.2.5. Given the algebraic linear system in Eqs. (5.2.1)-(5.2.2), introduce the
coefficient matrix A, the unknown vector x, and the source vector b, as follows,
⎡
⎤ ⎡ ⎤ ⎡ ⎤
A =
⎢
⎣
a11 · · · a1n
.
an1 · · · ann
.
⎥ ⎢
⎦ , x = ⎣
Then, the algebraic linear system can be written as
Ax = b.
x1
.
xn
⎥ ⎢
⎦ , b = ⎣
Proof of Theorem 5.2.5: From the definition of the matrix-vector product we have that
⎡
a11
⎢
Ax = ⎣ .
· · ·
⎤ ⎡ ⎤ ⎡
⎤
a1n x1 a11x1 + · · · + a1nxn
⎥ ⎢ ⎥ ⎢
⎥
. ⎦ ⎣ . ⎦ = ⎣ . ⎦ .
an1x1 + · · · + a1nxn
an1 · · · ann
Then, we conclude that
⎫
a11x1 + · · · + a1nxn = b1,
⎪⎬
.
⎪⎭
an1x1 + · · · + annxn = bn,
⇔
⎡
⎢
⎣
xn
a11x1 + · · · + a1nxn
.
an1x1 + · · · + a1nxn
⎤
⎥ ⎢
⎦ = ⎣
We introduce one last definition, which will be helpful in the next subsection.
⎡
b1
.
bn
b1
.
bn
⎥
⎦ .
⎤
⎥
⎦ ⇔ Ax = b.
Definition 5.2.6. The augmented matrix of Ax = b is the n × (n + 1) matrix [A|b].
The augmented matrix of an algebraic linear system contains the equation coefficients and
the sources. Therefore, the augmented matrix of a linear system contains the complete
information about the system.
⊳

G. NAGY – ODE July 2, 2013 157
Example 5.2.6: Find the augmented matrix of both the linear systems in Example 5.2.1.
Solution: The coefficient matrix and source vector of the first system imply that
2
A =
−1
−1
,
2
0
b =
3
⇒
2
[A|b] =
−1
−1
2
0
.
3
The coefficient matrix and source vector of the second system imply that
⎡
1
A = ⎣−3
0
2
1
1
⎤
1
3 ⎦ ,
−4
⎡ ⎤
1
b = ⎣ 24 ⎦
−1
⇒
⎡
1
[A|b] = ⎣−3
0
2
1
1
1
3
−4
⎤
1
24 ⎦ .
−1
Recall that the linear combination of two vectors is defined component-wise, that is, given
any numbers a, b ∈ R and any vectors x, y, their linear combination is the vector given by
⎡ ⎤
⎡ ⎤ ⎡ ⎤
ax1 + by1 x1 y1 ⎢ ⎥
⎢ ⎥ ⎢ ⎥
ax + by = ⎣ . ⎦ , where x = ⎣ . ⎦ , y = ⎣ . ⎦ .
axn + byn
With this definition of linear combination of vectors it is simple to see that the matrix-vector
product is a linear operation.
Theorem 5.2.7. The matrix-vector product is a linear operation, that is, given an n × n
matrix A, then for all n-vectors x, y and all numbers a, b ∈ R holds
xn
A(ax + by) = a Ax + b Ay. (5.2.3)
Proof of Theorem 5.2.7: Just write down the matrix-vector product in components,
⎡
a11
⎢
A(ax+by) = ⎣ .
· · ·
⎤ ⎡ ⎤ ⎡
⎤
a1n ax1 + by1 a11(ax1 + by1) + · · · + a1n(axn + byn)
⎥ ⎢ ⎥ ⎢
⎥
. ⎦ ⎣ . ⎦ = ⎣
.
⎦ .
am1 · · · amn axn + byn an1(ax1 + by1) + · · · + ann(axn + byn)
Expand the linear combinations on each component on the far right-hand side above and
re-order terms as follows,
⎡
⎤
A(ax + by) =
⎢
⎣
Separate the right-hand side above,
⎡
A(ax + by) = a
We then conclude that
⎢
⎣
a (a11x 1 + · · · + a1nxn) + b (a11y 1 + · · · + a1nyn)
.
a (an1x 1 + · · · + annxn) + b (an1y 1 + · · · + annyn)
(a11x1 + · · · + a1nxn)
.
(an1x1 + · · · + annxn)
⎤
⎡
⎥ ⎢
⎦ + b ⎣
A(ax + by) = a Ax + b Ay.
yn
⎥
⎦ .
(a11y1 + · · · + a1nyn)
.
(an1y1 + · · · + annyn)
This establishes the Theorem.
⎤
⎥
⎦ .
⊳

158 G. NAGY – ODE july 2, 2013
5.2.2. Gauss elimination operations. We now review three operations that can be performed
on an augmented matrix of a linear system. These operations change the augmented
matrix of the system but they do not change the solutions of the system. The Gauss elimination
operations were already known in China around 200 BC. We call them after Carl
Friedrich Gauss, since he made them very popular around 1810, when he used them to study
the orbit of the asteroid Pallas, giving a systematic method to solve a 6 × 6 algebraic linear
system.
Definition 5.2.8. The Gauss elimination operations are three operations on a matrix:
(i) Adding to one row a multiple of the another;
(ii) Interchanging two rows;
(iii) Multiplying a row by a non-zero number.
These operations are respectively represented by the symbols given in Fig. 23.
a
a = 0
Figure 23. A representation of the Gauss elimination operations.
As we said above, the Gauss elimination operations change the coefficients of the augmented
matrix of a system but do not change its solution. Two systems of linear equations having
the same solutions are called equivalent. It can be shown that there is an algorithm using
these operations that transforms any n × n linear system into an equivalent system where
the solutions are explicitly given.
Example 5.2.7: Find the solution to the 2 × 2 linear system given in Example 5.2.1 using
the Gauss elimination operations.
Solution: Consider the augmented matrix of the 2 × 2 linear system in Example (5.2.1),
and perform the following Gauss elimination operations,
2 −1
0 2 −1 0
−1 2 →
2 −1
3 −2 4 →
6 0 3
0 2
→
6 0
−1
1
0
→
2
2
0
0
1
2 1
→
2 0
0
1
1
2
⇔
x1 + 0 = 1
0 + x2 = 2
⇔
x1 = 1
x2 = 2
⊳
Example 5.2.8: Find the solution to the 3 × 3 linear system given in Example 5.2.1 using
the Gauss elimination operations
Solution: Consider the augmented matrix of the 3 × 3 linear system in Example 5.2.1 and
perform the following Gauss elimination operations,
⎡
⎤ ⎡
⎤ ⎡
1 2 1
1 1 2 1
1 1
⎣−3
1 3
24 ⎦ → ⎣0
7 6 27 ⎦
→ ⎣0
0 1 −4 −1 0 1 −4 −1 0
2
1
7
1
−4
6
⎤
1
−1⎦
,
27
⎡
1
⎣0
0
0
1
0
9
−4
34
⎤ ⎡
3 1
−1⎦
→ ⎣0
34 0
0
1
0
9
−4
1
⎤ ⎡
3 1
−1⎦
→ ⎣0
1 0
0
1
0
0
0
1
⎤
−6
3 ⎦
1
⇒
⎧
⎪⎨
x1 = −6,
x2 = 3,
⎪⎩
x3 = 1.
⊳

G. NAGY – ODE July 2, 2013 159
In the last augmented matrix on both Examples 5.2.7 and 5.2.8 the solution is given explicitly.
This is not always the case with every augmented matrix. A precise way to define
the final augmented matrix in the Gauss elimination method is captured in the notion of
echelon form and reduced echelon form of a matrix.
Definition 5.2.9. An m × n matrix is in echelon form iff the following conditions hold:
(i) The zero rows are located at the bottom rows of the matrix;
(ii) The first non-zero coefficient on a row is always to the right of the first non-zero
coefficient of the row above it.
The pivot coefficient is the first non-zero coefficient on every non-zero row in a matrix in
echelon form.
Example 5.2.9: The 6 × 8, 3 × 5 and 3 × 3 matrices given below are in echelon form, where
the ∗ means any non-zero number and pivots are highlighted.
⎡
⎤
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
⎢
⎢0
0 ∗ ∗ ∗ ∗ ∗ ∗⎥
⎡
⎤
⎥
⎢
⎢0
0 0 ∗ ∗ ∗ ∗ ∗⎥
∗ ∗ ∗ ∗ ∗
⎥
⎢
⎢0
0 0 0 0 0 ∗ ∗⎥
, ⎣0 0 ∗ ∗ ∗⎦
,
⎥
⎣0
0 0 0 0 0 0 0⎦
0 0 0 0 0
⎡
∗
⎣0 0
∗
∗
0
⎤
∗
∗⎦
.
∗
0 0 0 0 0 0 0 0
⊳
Example 5.2.10: The following matrices are in echelon form, with pivot highlighted.
⎡ ⎤
2 1 1
1 3 2 3 2
,
, ⎣0 3 4⎦
.
0 1 0 4 −2
0 0 0
Definition 5.2.10. An m × n matrix is in reduced echelon form iff the matrix is in
echelon form and the following two conditions hold:
(i) The pivot coefficient is equal to 1;
(ii) The pivot coefficient is the only non-zero coefficient in that column.
We denote by EA a reduced echelon form of a matrix A.
Example 5.2.11: The 6×8, 3×5 and 3×3 matrices given below are in echelon form, where
the ∗ means any non-zero number and pivots are highlighted.
⎡
⎤
1 ∗ 0 0 ∗ ∗ 0 ∗
⎢
⎢0
0 1 0 ∗ ∗ 0 ∗⎥
⎡
⎤
⎥
⎢
⎢0
0 0 1 ∗ ∗ 0 ∗⎥
1 ∗ 0 ∗ ∗
⎥
⎢
⎢0
0 0 0 0 0 1 ∗⎥
, ⎣0
0 1 ∗ ∗⎦
,
⎥
⎣0
0 0 0 0 0 0 0⎦
0 0 0 0 0
⎡
1
⎣0
0
0
1
0
⎤
0
0⎦
.
1
0 0 0 0 0 0 0 0
⊳
Example 5.2.12: And the following matrices are not only in echelon form but also in
reduced echelon form; again, pivot coefficients are highlighted.
⎡ ⎤
1 0 0
1 0 1 0 4
,
, ⎣0
1 0⎦
.
0 1 0 1 5
0 0 0
⊳
Summarizing, the Gauss elimination operations can transform any matrix into reduced
echelon form. Once the augmented matrix of a linear system is written in reduced echelon
form, it is not difficult to decide whether the system has solutions or not.
⊳

160 G. NAGY – ODE july 2, 2013
Example 5.2.13: Use Gauss operations to find the solution of the linear system
2x1 − x2 = 0,
− 1
2 x1 + 1
4 x2 = − 1
4 .
Solution: We find the system augmented matrix and perform appropriate Gauss elimination
operations,
2 −1 0
− 1
2 −1 0
1 1 →
2 −1 0
2 4 − 4 −2 1 →
−1 0 0 1
From the last augmented matrix above we see that the original linear system has the same
solutions as the linear system given by
2x1 − x2 = 0,
0 = 1.
Since the latter system has no solutions, the original system has no solutions. ⊳
The situation shown in Example 5.2.13 is true in general. If the augmented matrix [A|b] of
an algebraic linear system is transformed by Gauss operations into the augmented matrix
[ Ã|˜ b] having a row of the form [0, · · · , 0|1], then the original algebraic linear system Ax = b
has no solution.
Example 5.2.14: Find all vectors b such that the system Ax = b has solutions, where
⎡
1
A = ⎣−1 −2
1
⎤
3
−2⎦
,
⎡
b = ⎣
2 −1 3
b1
⎤
b2⎦
.
Solution: We do not need to write down the algebraic linear system, we only need its
augmented matrix,
⎡
1
[A|b] = ⎣−1
2
−2
1
−1
3
−2
3
⎤ ⎡
b1 1
b2⎦
→ ⎣0
b3 2
−2
−1
−1
3
1
3
⎤
b1
b1 + b2⎦
→
b3
⎡
1
⎣0
0
−2
1
3
3
−1
−3
⎤ ⎡
b1 1
−b1 − b2⎦
→ ⎣0
b3 − 2b1 0
−2
1
0
3
−1
0
⎤
b1
−b1 − b2 ⎦ .
b3 + b1 + 3b2
Therefore, the linear system Ax = b has solutions
⇔ the source vector satisfies the equation
holds b1 + 3b2 + b3 = 0.
That is, every source vector b that lie on the
plane normal to the vector n is a source vector
such that the linear system Ax = b has
solution, where
⎡ ⎤
1
n = ⎣3⎦
.
1
⊳
b3
b 1
b
3
b
n
b
2

G. NAGY – ODE July 2, 2013 161
5.2.3. Linearly dependence. We generalize the idea of two vectors lying on the same line,
and three vectors lying on the same plane, to an arbitrary number of vectors.
Definition 5.2.11. A set of vectors {v 1, · · · , vk}, with k 1 is called linearly dependent
iff there exists constants c1, · · · , ck, with at least one of them non-zero, such that
c1 v1 + · · · + ck vk = 0. (5.2.4)
The set of vectors is called linearly independent iff it is not linearly dependent, that is,
the only constants c1, · · · , ck that satisfy Eq. (5.2.4) are given by c1 = · · · = ck = 0.
In other words, a set of vectors is linearly dependent iff one of the vectors is a linear combination
of the other vectors. When this is not possible, the set is called linearly independent.
Example 5.2.15: Show that the following set of vectors is linearly dependent,
⎡
⎣ 1
⎤ ⎡
2⎦
, ⎣
3
3
⎤ ⎡
2⎦
, ⎣
1
−1
⎤
2 ⎦
5
,
and express one of the vectors as a linear combination of the other two.
Solution: We need to find constants c1, c2, and c3 solutions of the equation
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
1 3 −1 0 1 3 −1 c1 0
⎣2⎦
c1 + ⎣2⎦
c2 + ⎣ 2 ⎦ c3 = ⎣0⎦
⇔ ⎣2
2 2 ⎦ ⎣c
⎦ 2 + ⎣0⎦
.
3 1 5 0 3 1 5 c3 0
The solution to this linear system can be obtained with Gauss elimination operations,
⎡ ⎤ ⎡
⎤ ⎡ ⎤ ⎡ ⎤ ⎧
1 3 −1 1 3 −1 1 3 −1 1 0 2 ⎪⎨
c1 = −2c3, ⎣2 2 2 ⎦ → ⎣0
−4 4 ⎦ → ⎣0
1 −1⎦
→ ⎣0
1 −1⎦
⇒ c2 = c3, ⎪⎩
3 1 5 0 −8 8 0 1 −1 0 0 0
c3 = free.
Since there are non-zero constants c1, c2, c3 solutions to the linear system above, the vectors
are linearly dependent. Choosing c3 = −1 we obtain the third vector as a linear combination
of the other two vectors,
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
1 3 −1 0
2 ⎣2⎦
− ⎣2⎦
− ⎣ 2 ⎦ = ⎣0⎦
⇔
3 1 5 0
−1 1 3
⎣ 2 ⎦ = 2 ⎣2⎦
− ⎣2⎦
.
5 3 1
⊳

162 G. NAGY – ODE july 2, 2013
5.2.4. Exercises.
5.2.1.- . 5.2.2.- .

G. NAGY – ODE July 2, 2013 163
5.3. Matrix algebra
The matrix-vector product introduced in Section 5.2 implies that an n × n matrix A is a
function A : R n → R n . This idea leads to introduce matrix operations, like the operations
introduced for functions f : R → R. These operations include linear combinations of
matrices; the composition of matrices, also called matrix multiplication; the inverse of a
square matrix; and the determinant of a square matrix. These operations will be needed in
later Sections when we study systems of differential equations.
5.3.1. A matrix is a function. The matrix-vector product leads to the interpretation
that an n × n matrix A is a function. If we denote by R n the space of all n-vectors, we see
that the matrix-vector product associates to the n-vector x the unique n-vector y = Ax.
Therefore the matrix A determines a function A : R n → R n .
Example 5.3.1: Describe the action on R2 of the function given by the 2 × 2 matrix
0 1
A = . (5.3.1)
1 0
Solution: The action of this matrix on an arbitrary element x ∈ R2 is given below,
0 1 x1 x2
Ax =
= .
1 0
Therefore, this matrix interchanges the components x 1 and x 2 of the vector x. It can be
seen in the first picture in Fig. 24 that this action can be interpreted as a reflection on the
plane along the line x1 = x2. ⊳
x
2
Ax
x = x
1
2
x
x
1
Figure 24. Geometrical meaning of the function determined by the matrix
in Eq. (5.3.1) and the matrix in Eq. (5.3.2), respectively.
Example 5.3.2: Describe the action on R2 of the function given by the 2 × 2 matrix
0 −1
A = . (5.3.2)
1 0
Solution: The action of this matrix on an arbitrary element x ∈ R2 is given below,
0 −1 x1 −x2
Ax =
= .
1 0
In order to understand the action of this matrix, we give the following particular cases:
0 −1 1 0 0 −1 1 −1 0 −1 −1 0
= ,
= ,
= .
1 0 0 1 1 0 1 1 1 0 0 −1
x2
x2
Ay
z
x1
x1
x
2
Ax
Az
x
y
x
1

164 G. NAGY – ODE july 2, 2013
These cases are plotted in the second figure on Fig. 24, and the vectors are called x, y and
z, respectively. We therefore conclude that this matrix produces a ninety degree counterclockwise
rotation of the plane. ⊳
An example of a scalar-valued function is f : R → R. We have seen here that an n×n matrix
A is a function A : R n → R n . Therefore, one can think an n × n matrix as a generalization
of the concept of function from R to R n , for any positive integer n. It is well-known how to
define several operations on scalar-valued functions, like linear combinations, compositions,
and the inverse function. Therefore, it is reasonable to ask if these operation on scalarvalued
functions can be generalized as well to matrices. The answer is yes, and the study
of these and other operations is the subject of the rest of this Section.
5.3.2. Matrix operations. The linear combination of matrices refers to the addition of two
matrices and the multiplication of a matrix by scalar. Linear combinations of matrices are
defined component by component. For this reason we introduce the component notation
for matrices and vectors. We denote an m × n matrix by A = [Aij], where Aij are the
components of matrix A, with i = 1, · · · , m and j = 1, · · · , n. Analogously, an n-vector is
denoted by v = [vj], where vj are the components of the vector v. We also introduce the
notation F = {R, C}, that is, the set F can be the real numbers or the complex numbers.
Definition 5.3.1. Let A = [Aij] and B = [Bij] be m × n matrices in F m,n and a, b be
numbers in F. The linear combination of A and B is also and m × n matrix in F m,n ,
denoted as a A + b B, and given by
a A + b B = [a Aij + b Bij].
The particular case where a = b = 1 corresponds to the addition of two matrices, and the
particular case of b = 0 corresponds to the multiplication of a matrix by a number, that is,
A + B = [Aij + Bij], a A = [a Aij].
1 2 2 3
Example 5.3.3: Find the A + B, where A = , B = .
3 4 5 1
Solution: The addition of two equal size matrices is performed component-wise:
1 2 2 3 (1 + 2) (2 + 3) 3 5
A + B = + =
= .
3 4 5 1 (3 + 5) (4 + 1) 8 5
1 2 1 2 3
Example 5.3.4: Find the A + B, where A = , B = .
3 4 4 5 6
Solution: The matrices have different sizes, so their addition is not defined. ⊳
1
Example 5.3.5: Find 2A and A/3, where A =
2
3
4
5
.
6
Solution: The multiplication of a matrix by a number is done component-wise, therefore
1
2A = 2
2
3
4
5 2
=
6 4
6
8
10
,
12
A 1 1
=
3 3 2
3
4
⎡1
5 ⎢3
= ⎢
6 ⎣
2
3
1
4
3
5⎤
3 ⎥
⎦
2
.
⊳
Since matrices are generalizations of scalar-valued functions, one can define operations
on matrices that, unlike linear combinations, have no analogs on scalar-valued functions.
One of such operations is the transpose of a matrix, which is a new matrix with the rows
and columns interchanged.
⊳

G. NAGY – ODE July 2, 2013 165
Definition 5.3.2. The transpose of a matrix A = [Aij] ∈ Fm,n is the matrix denoted as
AT = (AT
n,m T
)kl ∈ F , with its components given by A
= Alk.
kl
1 3 5
Example 5.3.6: Find the transpose of the 2 × 3 matrix A = .
2 4 6
Solution: Matrix A has components Aij with i = 1, 2 and j = 1, 2, 3. Therefore, its
transpose has components (A T )ji = Aij, that is, A T has three rows and two columns,
A T ⎡
1
⎤
2
= ⎣3
4⎦
.
5 6
If a matrix has complex-valued coefficients, then the conjugate of a matrix can be defined
as the conjugate of each component.
Definition 5.3.3. The complex conjugate of a matrix A = [Aij] ∈ Fm,n is the matrix
A =
m,n
Aij ∈ F .
Example 5.3.7: A matrix A and its conjugate is given below,
1
A =
−i
2 + i
,
3 − 4i
⇔
1
A =
i
2 − i
.
3 + 4i
Example 5.3.8: A matrix A has real coefficients iff A = A; It has purely imaginary coefficients
iff A = −A. Here are examples of these two situations:
1 2
1 2
A =
⇒ A = = A;
3 4
3 4
i 2i
−i −2i
A =
⇒ A =
= −A.
3i 4i
−3i −4i
⊳
Definition 5.3.4. The adjoint of a matrix A ∈ Fm,n is the matrix A∗ = A T
∈ Fn,m .
Since A T = (AT ), the order of the operations does not change the result, that is why there
is no parenthesis in the definition of A∗ .
Example 5.3.9: A matrix A and its adjoint is given below,
1
A =
−i
2 + i
,
3 − 4i
⇔ A ∗
1
=
2 − i
i
.
3 + 4i
The transpose, conjugate and adjoint operations are useful to specify certain classes of
matrices with particular symmetries. Here we introduce few of these classes.
Definition 5.3.5. An n × n matrix A is called:
(a) symmetric iff holds A = A T ;
(b) skew-symmetric iff holds A = −A T ;
(c) Hermitian iff holds A = A ∗ ;
(d) skew-Hermitian iff holds A = −A ∗ .
⊳
⊳
⊳

166 G. NAGY – ODE july 2, 2013
Example 5.3.10: We present examples of each of the classes introduced in Def. 5.3.5.
Part (a): Matrices A and B are symmetric. Notice that A is also Hermitian, while B is
not Hermitian,
⎡
1 2
⎤
3
A = ⎣2
7 4⎦
= A
3 4 8
T ⎡
1 2 + 3i
⎤
3
, B = ⎣2
+ 3i 7 4i⎦
= B
3 4i 8
T .
Part (b): Matrix C is skew-symmetric,
⎡
0
C = ⎣ 2
−2
0
⎤
3
−4⎦
⇒ C
−3 4 0
T ⎡
0
= ⎣−2 2
0
⎤
−3
4 ⎦ = −C.
3 −4 0
Notice that the diagonal elements in a skew-symmetric matrix must vanish, since Cij = −Cji
in the case i = j means Cii = −Cii, that is, Cii = 0.
Part (c): Matrix D is Hermitian but is not symmetric:
⎡
1 2 + i 3
⎤
D = ⎣2
− i 7 4 + i⎦
⇒ D
3 4 − i 8
T ⎡
1 2 − i 3
⎤
= ⎣2
+ i 7 4 − i⎦
= D,
3 4 + i 8
however,
D ∗ = D T
⎡
1
= ⎣2 − i
2 + i
7
⎤
3
4 + i⎦
= D.
3 4 − i 8
Notice that the diagonal elements in a Hermitian matrix must be real numbers, since the
condition Aij = Aji in the case i = j implies Aii = Aii, that is, 2iIm(Aii) = Aii − Aii = 0.
We can also verify what we said in part (a), matrix A is Hermitian since A∗ = A T
= AT Part (d): The following matrix E is skew-Hermitian:
= A.
⎡
i
E = ⎣−2 + i
2 + i
7i
⎤
−3
4 + i⎦
⇒ E
3 −4 + i 8i
T ⎡
i
= ⎣2 + i
−2 + i
7i
⎤
3
−4 + i⎦
−3 4 + i 8i
therefore,
E ∗ = E T
⎡
−i
⎣2
− i
−2 − i
−7i
⎤
3
−4 − i⎦
= −E.
−3 4 − i −8i
A skew-Hermitian matrix has purely imaginary elements in its diagonal, and the off diagonal
elements have skew-symmetric real parts with symmetric imaginary parts. ⊳
The trace of a square matrix is a number, the sum of all the diagonal elements of the matrix.
Definition 5.3.6. The trace of a square matrix A =
n,n
Aij ∈ F , denoted as tr (A) ∈ F,
is the sum of its diagonal elements, that is, the scalar given by tr (A) = A11 + · · · + Ann.
⎡
1
Example 5.3.11: Find the trace of the matrix A = ⎣4 2
5
⎤
3
6⎦.
7 8 9
Solution: We only have to add up the diagonal elements:
tr (A) = 1 + 5 + 9 ⇒ tr (A) = 15.
⊳

G. NAGY – ODE July 2, 2013 167
The operation of matrix multiplication originates in the composition of functions. We
call it matrix multiplication instead of matrix composition because it reduces to the multiplication
of real numbers in the case of 1 × 1 real matrices. Unlike the multiplication of
real numbers, the product of general matrices is not commutative, that is, AB = BA in
the general case. This property reflects the fact that the composition of two functions is a
non-commutative operation.
Definition 5.3.7. The matrix multiplication of the m × n matrix A = [Aij] and the
n × ℓ matrix B = [Bjk], where i = 1, · · · , m, j = 1, · · · , n and k = 1, · · · , ℓ, is the m × ℓ
matrix AB given by
n
(AB)ik = AijBjk. (5.3.3)
j=1
The product is not defined for two arbitrary matrices, since the size of the matrices is
important: The numbers of columns in the first matrix must match the numbers of rows in
the second matrix.
A times B defines AB
m × n n × ℓ m × ℓ
2
Example 5.3.12: Compute AB, where A =
−1
−1
and B =
2
3 0
.
2 −1
Solution: The component (AB)11 = 4 is obtained from the first row in matrix A and the
first column in matrix B as follows:
2 −1 3 0 4
=
−1 2 2 −1 1
1
,
−2
(2)(3) + (−1)(2) = 4;
The component (AB)12 = −1 is obtained as follows:
2 −1 3 0 4 1
= , (2)(0) + (−1)(1) = −1;
−1 2 2 −1 1 −2
The component (AB)21 = 1 is obtained as follows:
2 −1 3 0 4 1
= , (−1)(3) + (2)(2) = 1;
−1 2 2 −1 1 −2
And finally the component (AB)22 = −2 is obtained as follows:
2 −1 3 0 4 1
= , (−1)(0) + (2)(−1) = −2.
−1 2 2 −1 1 −2
Example 5.3.13: Compute BA, where A =
Solution: We find that BA =
Example 5.3.14: Compute AB and BA, where A =
Solution: The product AB is
4 3 1 2 3
AB =
2 1 4 5 6
2 −1
and B =
−1 2
3 0
.
2 −1
6 −3
. Notice that in this case AB = BA. ⊳
5 −4
⇒ AB =
4 3
and B =
2 1
16 23 30
.
6 9 12
1 2 3
.
4 5 6
The product BA is not possible. ⊳
⊳

168 G. NAGY – ODE july 2, 2013
5.3.3. The inverse matrix. We now introduce the concept of the inverse of a square
matrix. Not every square matrix is invertible. The inverse of a matrix is useful to compute
solutions to linear systems of algebraic equations.
Definition 5.3.8. The matrix In ∈ F n,n is the identity matrix iff Inx = x for all x ∈ F n .
It is simple to see that the components of the identity matrix are given by
Iii = 1
The cases n = 2, 3 are given by
In = [Iij] with
I2 =
Iij = 0 i = j.
⎡ ⎤
1 0 0
1 0
, I3 = ⎣0 1 0⎦
.
0 1
0 0 1
Definition 5.3.9. A matrix A ∈ F n,n is called invertible iff there exists a matrix, denoted
as A −1 , such that A −1 A = In, and A A −1 = In.
Example 5.3.15: Verify that the matrix and its inverse are given by
2 2
A = , A
1 3
−1 = 1
3 −2
.
4 −1 2
Solution: We have to compute the products,
A A −1
2 2 1 3 −2
=
=
1 3 4 −1 2
1
4 0
4 0 4
⇒ A A −1 = I2.
It is simple to check that the equation A −1 A = I2 also holds. ⊳
Theorem 5.3.10. Given a 2 × 2 matrix A introduce the number ∆ as follows,
a b
A = , ∆ = ad − bc.
c d
The matrix A is invertible iff ∆ = 0. Furthermore, if A is invertible, its inverse is given by
A −1 = 1
∆
d −b
. (5.3.4)
−c a
The number ∆ is called the determinant of A, since it is the number that determines whether
A is invertible or not.
2 2
Example 5.3.16: Compute the inverse of matrix A = , given in Example 5.3.15.
1 3
Solution: Following Theorem 5.3.10 we first compute ∆ = 6 − 4 = 4. Since ∆ = 0, then
A−1 exists and it is given by
A −1 = 1
3
4 −1
−2
.
2
⊳
1
Example 5.3.17: Compute the inverse of matrix A =
3
2
.
6
Solution: Following Theorem 5.3.10 we first compute ∆ = 6 − 6 = 0. Since ∆ = 0, then
matrix A is not invertible. ⊳

G. NAGY – ODE July 2, 2013 169
Gauss operations can be used to compute the inverse of a matrix. The reason for this is
simple to understand in the case of 2 × 2 matrices, as can be seen in the following Example.
Example 5.3.18: Given any 2 × 2 matrix A, find its inverse matrix, A −1 , or show that the
inverse does not exist.
Solution: If the inverse matrix, A −1 exists, then denote it as A −1 = [x1, x2]. The equation
A(A −1 ) = I2 is then equivalent to A [x1, x2] =
two algebraic linear systems,
A x1 =
1 0
. This equation is equivalent to solving
0 1
1
, A x2 =
0
0
.
1
Here is where we can use Gauss elimination operations. We use them on both systems
1
0
A , A .
0
1
However, we can solve both systems at the same time if we do Gauss operations on the
bigger augmented matrix
1
A
0
0
.
1
Now, perform Gauss operations until we obtain the reduced echelon form for [A|I2]. Then
we can have two different types of results:
• If there is no line of the form [0, 0|∗, ∗] with any of the star coefficients non-zero,
then matrix A is invertible and the solution vectors x1, x2 form the columns of the
inverse matrix, that is, A−1 = [x1, x2].
• If there is a line of the form [0, 0|∗, ∗] with any of the star coefficients non-zero, then
matrix A is not invertible. ⊳
2 2
Example 5.3.19: Use Gauss operations to find the inverse of A = .
1 3
Solution: As we said in the Example above, perform Gauss operation on the augmented
matrix [A|I2] until the reduced echelon form is obtained, that is,
2 2
1 0 1 3 0 1
1 3 →
1 3
0 1 2 2 → 0
1 0 0 −4 1
1 3
0 1 1 0 3 1
0 1 1 1 → 4 − 2
− 4 2 0 1 1 1 − 4 2
1
→
−2
That is, matrix A is invertible and the inverse is
A −1
3 1
= 4 − 2
− 1
1 ⇔ A
4 2
−1 = 1
3
4 −1
−2
.
2
⎡
1 2
⎤
3
⊳
Example 5.3.20: Use Gauss operations to find the inverse of A = ⎣2
5 7⎦.
3 7 9
Solution: We perform Gauss operations on the augmented matrix [A|I3] until we obtain
its reduced echelon form, that is,
⎡
1 2 3
1
⎣2
5 7
0
3 7 9 0
0
1
0
⎤ ⎡
0 1
0⎦
→ ⎣0
1 0
2
1
1
3
1
0
1
−2
−3
0
1
0
⎤
0
0⎦
→
1

170 G. NAGY – ODE july 2, 2013
⎡
1
⎣0
0
0
1
0
⎤ ⎡
1
5 −2 0 1
1
−2 1 0⎦
→ ⎣
−1 −1 −1 1
0 1 0 1 1
5 −2 0
0 0 1
−2 1 0
⎡
1
⎣0
0
0
1
0
⎤ ⎡
1
5 −2 0 1
1 −2 1 0 ⎦
→ ⎣0
1 1 1 −1 0
0
1
0
0
0
1
1
4
−3
1
1
−3
0
1
⎤
⎦
−1
⎤
1
1 ⎦
−1
We conclude that matrix A is invertible and
A −1 ⎡
4
= ⎣−3 −3
0
⎤
1
1 ⎦ .
1 1 −1
5.3.4. Determinants. A determinant is a scalar computed form a square matrix that gives
important information about the matrix, for example if the matrix is invertible or not. We
now review the definition and properties of the determinant of 2 × 2 and 3 × 3 matrices.
a11 a12
Definition 5.3.11. The determinant of a 2 × 2 matrix A =
is given by
det(A) =
a11 a12
a21 a22
The determinant of a 3 × 3 matrix A = ⎣
det(A) =
a11 a12 a13
a21 a22 a23
a31 a32 a33
= a11
= a11a22 − a12a21.
⎡
⎤
a22 a23
a32 a33
a11 a12 a13
a21 a22 a23
a31 a32 a33
− a12
a21 a22
⎦ is given by
a21 a23
a31 a33
+ a13
a21 a22
a31 a32
Example 5.3.21: The following three examples show that the determinant can be a negative,
zero or positive number.
1
2
3 4
= 4 − 6 = −2,
2
3
1
4
= 8 − 3 = 5,
1
2
2
4
= 4 − 4 = 0.
The following is an example shows how to compute the determinant of a 3 × 3 matrix,
1
3 −1
2
1 1
= (1) 1
1
3
2 1 2 1
− 3 2
1
3 1
+ (−1) 2
1
3 2
= (1 − 2) − 3 (2 − 3) − (4 − 3)
= −1 + 3 − 1
= 1. ⊳
The absolute value of the determinant of a 2 × 2 matrix A = [a1, a2] has a geometrical
meaning: It is the area of the parallelogram whose sides are given by a1 and bia2, that is, by
the columns of the matrix A; see Fig. 25. Analogously, the absolute value of the determinant
of a 3 × 3 matrix A = [a 1, a 2, a 3] also has a geometrical meaning: It is the volume of the
parallelepiped whose sides are given by a 1, a 2 and a 3, that is, by the columns of the matrix
A; see Fig. 25.
The determinant of an n × n matrix A can be defined generalizing the properties that
areas of parallelogram have in two dimensions and volumes of parallelepipeds have in three
dimensions. One of these properties is the following: if one of the column vectors of the
.
⊳

0
a 1
R
2
G. NAGY – ODE July 2, 2013 171
a 2
Figure 25. Geometrical meaning of the determinant.
matrix A is a linear combination of the others, then the figure determined by these column
vectors is not n-dimensional but (n−1)-dimensional, so its volume must vanish. We highlight
this property of the determinant of n × n matrices in the following result.
Theorem 5.3.12. The set of n-vectors {v 1, · · · , vn}, with n 1, is linearly dependent iff
a 1
det[v1, · · · , vn] = 0.
Example 5.3.22: Show whether the set of vectors below linearly independent,
⎡
⎣ 1
⎤ ⎡
2⎦
, ⎣
3
3
⎤ ⎡
2⎦
, ⎣
1
−3
⎤
2 ⎦
7
.
The determinant of the matrix whose column vectors are the vectors above is given by
1
3 −3
2
2 2
= (1) (14 − 2) − 3 (14 − 6) + (−3) (2 − 6) = 12 − 24 + 12 = 0.
3
1 7
Therefore, the set of vectors above is linearly dependent. ⊳
The determinant of a square matrix also determines whether the matrix is invertible or not.
Theorem 5.3.13. An n × n matrix A is invertible iff holds det(A) = 0.
⎡
1 2
⎤
3
Example 5.3.23: Is matrix A = ⎣2
5 7⎦
invertible?
3 7 9
Solution: We only need to compute the determinant of A.
1
2 3
det(A) =
2
5 7
= (1) 5
7
3
7 9
7 9
− (2) 2
7
3 9
+ (3) 2
5
3 7
.
Using the definition of determinant for 2 × 2 matrices we obtain
det(A) = (45 − 49) − 2(18 − 21) + 3(14 − 15) = −4 + 6 − 3.
Since det(A) = −1, that is, non-zero, matrix A is invertible. ⊳
R 3
a 2
a
3

172 G. NAGY – ODE july 2, 2013
5.3.5. Exercises.
5.3.1.- . 5.3.2.- .

G. NAGY – ODE July 2, 2013 173
5.4. Linear differential Systems
In Section 5.1 we introduced differential systems and the particular case of linear differential
systems. We also defined the initial value problem for differential systems and stated
a version of the Picard-Lindelöf Theorem. This result says that the initial value problem for
certain differential systems has a unique solution. In this Section we focus on linear differential
systems. We use matrix-vector notation to simplify the presentation. We define several
useful concepts like fundamental solutions, general solutions, fundamental matrix, and the
Wronskian. We also describe few properties shared by solutions to linear differential systems.
The superposition property of solutions to homogeneous linear differential systems is
one example. The generalization of Abel’s Theorem from a single linear differential equation
to a linear differential system is another example.
5.4.1. Matrix-vector notation. In Section 5.1, Definition 5.1.2 we introduced an n × n
first order system of linear differential equations. We said that one has to fix both the
coefficient and source functions aij, gi : [t1, t2] → R, where i, j = 1, · · · , n, and then one has
to find n functions xj : [t1, t2] → R solutions of the n equations
x ′ 1 = a11(t) x1 + · · · + a1n(t) xn + g1(t),
.
x ′ n = an1(t) x 1 + · · · + ann(t) xn + gn(t).
We also said that the system is called homogeneous iff the source functions satisfy that
g1 = · · · = gn = 0, and the system is called of constant coefficients iff all functions aij are
constants. Matrix-vector notation is very helpful to express linear differential systems in a
compact way. Since the coefficients aij, the sources gi, and the unknowns xj are functions,
we need to introduce a matrix-valued function and two vector-valued functions,
⎡
⎤
⎡ ⎤
⎡ ⎤
a11(t) · · · a1n(t)
g1(t)
x1(t)
⎢
⎥
⎢ ⎥
⎢ ⎥
A(t) = ⎣ .
. ⎦ , g(t) = ⎣ . ⎦ , x(t) = ⎣ . ⎦ .
an1(t) · · · ann(t)
gn(t)
xn(t)
Define the derivative of a vector-valued function x : [t1, t2] → Rn component-wise, that is,
x ′ ⎡
x
⎢
(t) = ⎣
′ 1(t)
.
x ′ ⎤
⎥
⎦ . (5.4.1)
n(t)
Then, the n × n system of linear differential equations given above can be written in a
compact was as follows,
x ′ = A(t) x + g(t).
Example 5.4.1: Use matrix notation to write down the 2 × 2 system given by
x ′ 1 = x1 − x2,
x ′ 2 = −x 1 + x 2.
Solution: In this case, the matrix of coefficients and the unknown vector have the form
1
A =
−1
−1
,
1
x1(t)
x(t) = .
x2(t)

174 G. NAGY – ODE july 2, 2013
This is an homogeneous system, so the source vector g = 0. The differential equation can
be written as follows,
x ′ 1 = x1 − x2
x ′
′ x 1 ⇔
2 = −x1 + x x 2
′
1 −1 x1
=
⇔ x
2 −1 1 x2
′ = Ax.
Example 5.4.2: Find the explicit expression for the linear system x ′ = Ax + b, where
1
A =
3
3
,
1
t e
g(t) = ,
x1
x = .
Solution: The 2 × 2 linear system is given by
′ x 1
x ′
t
1 3 x1 e
=
+
2 3 1 x2 2e3t
2e 3t
x2
, ⇔ x′ 1 = x1 + 3x2 + e t ,
x ′ 2 = 3x1 + x2 + 2e 3t .
Example 5.4.3: Find and express in matrix form the first order reduction to the system
y ′′ + 2y ′ + 3y = sin(5t).
Solution: We then introduce the functions x 1 = y and x 2 = y ′ . Therefore, x 1 and x 2
satisfy the 2 × 2 differential linear system,
x ′ 1 = x 2,
x ′ 2 = −3x1 − 2x2 + sin(5t).
Introduce the matrix, source vector and unknown vector,
0
A =
−3
1
,
−2
0
g(t) = ,
sin(5t)
x1(t)
x(t) = ,
x2(t)
then the system above can be written as follows,
′ x 1
x ′
0 1 x1 0
=
+
⇔ x
2 −3 −2 x2 sin(5t)
′ = Ax + g(t).
⊳
Example 5.4.4: Show that the vector-valued functions x (1)
2
= e
1
2t and x (2)
1
= e
2
−t
are solutions to the 2 × 2 linear system x ′
3 −2
= Ax, where A = .
2 −2
Solution: We start computing the right-hand side in the differential equation above for
the function x (1) , that is,
Ax (1)
3 −2 2
=
e
2 −2 1
2t
4
= e
2
2t
2
= 2 e
1
2t
2
= 2e
1
2t
=
so we conclude that x (1) ′ = Ax (1) . Analogously,
Ax (2) =
3
−2 1
2 −2 2
e −t =
−1
−2
e −t = −
1
2
e −t =
1
−e
2
−t
=
2
e
1
2t ′
= x (1)′ ,
⊳
⊳
1
e
2
−t ′
= x (2) ′ ,
so we conclude that x (2) ′ = Ax (2) . ⊳

G. NAGY – ODE July 2, 2013 175
5.4.2. Homogeneous systems. May be the most important property satisfied by solutions
to linear homogeneous differential system is the superposition property. Given two solutions
of the homogeneous system, their linear combination is also a solution to that system. This
property will play a crucial role to compute solutions to linear systems.
Theorem 5.4.1 (Superposition). If each vector-valued function x (1) , x (2) : [t 1, t 2] → R n
is solution of the homogeneous linear system x ′ = A(t) x, then so is the linear combination
ax (1) + bx (2) , for all a, b ∈ R.
This Theorem contains the particular case a = b = 1. So, if x (1) and x (2) are solutions of
an homogeneous linear system, so is x (1) + x (2) . Another particular case is b = 0 and a
arbitrary. So, if x (1) is a solution of an homogeneous linear system, so is ax (1) .
Proof of Theorem 5.4.1: The vector-valued functions x (1) and x (2) are solutions to the
homogeneous linear system, that is,
x (1) ′ = Ax (1) , x (2) ′ = Ax (2) . (5.4.2)
It is simple to see that the derivative of a vector-valued function, as defined in Eq. (5.4.1),
is a linear operation, meaning that for all a, b ∈ R holds,
ax (1) + bx (2) ′ = a x (1) ′ + b x (2) ′ .
Replacing Eq. (5.4.2) above we get
ax (1) + bx (2) ′ = a Ax (1) + b Ax (2) .
It was shown in Theorem 5.2.7 that the matrix-vector product is a linear operation, that is,
it satisfies Eq. (5.2.3), therefore, a Ax (1) + b Ax (2) = A ax (1) + bx (2) . We then obtain,
ax (1) + bx (2) ′ = A ax (1) + bx (2) ,
showing that the vector ax (1) + bx (2) is solution of the homogeneous equation too. This
establishes the Theorem.
Example 5.4.5: Verify that the functions x (1)
1
= e
1
−2t and x (2)
−1
= e
1
4t are solutions
to the homogeneous linear system
x ′ = Ax, A =
Verify that x (1) + x (2) is also a solution.
1 −3
.
−3 1
Solution: The function x (1) is solution to the differential equation, since
x (1) ′
1
= −2 e
1
−2t , Ax (1)
1 −3 1
=
e
−3 1 1
−2t
−2
= e
−2
−2t
1
= −2 e
1
−2t .
We then conclude that x (1) ′ = Ax (1) . Analogously, the function x (2) is solution to the
differential equation, since
x (2) ′
−1
= 4 e
1
4t , Ax (2) =
1 −3
−3 1
−1
1
e 4t =
−4
e
4
4t
−1
= 4 e
1
4t .
We then conclude that x (2) ′ = Ax (2) . To show that x (1) + x (2) is also a solution we could
use the linearity of the matrix-vector product, as we did in the proof of the Theorem 5.4.1.
Here we choose the straightforward, although more obscure, calculation: On the one hand,
x (1) + x (2)
−2t 4t e − e
=
⇒ x (1) + x (2)
−2t 4t
′ −2e − 4e
=
.
e −2t + e 4t
−2e −2t + 4e 4t

176 G. NAGY – ODE july 2, 2013
On the other hand,
that is,
A x (1) + x (2) =
1 −3
−3 1
e −2t − e 4t
e −2t + e 4t
A x (1) + x (2) =
=
e −2t − e 4t − 3e −2t − 3e 4t
−2e −2t − 4e 4t
−3e −2t + 3e 4t + e −2t + e 4t
−2e −2t + 4e 4t
We conclude that x (1) + x (2) ′ = A x (1) + x (2) . ⊳
Given n solutions x (1) , · · · , x (n) to an n × n linear homogeneous differential system, one
can always construct a matrix having these solutions as columns, that is, x (1) , · · · , x (n) .
This solution matrix is an n×n matrix for every value of the parameter t, so we can compute
several things with such a matrix. For example we can compute its determinant. Since this
determinant has important applications, we give it a name.
Definition 5.4.2. Given n vector-valued functions x (1) , · · · , x (n) solutions to an n×n linear
homogeneous system x ′ = A(t) x, we call the Wronskian of these solutions to the function
w = det x (1) , · · · , x (n) .
Suppose that the linear system in Definition 5.4.2 is a first order reduction of a second order
linear homogeneous equation, y ′′ + a1y ′ + a0y = 0, studied in Sect. 2.1. That is, suppose
that the linear system is
x ′ 1 = x 2, x ′ 2 = −a 0x 1 − a 1x 2.
In this case, the Wronskian defined here coincides with the definition given in Sect. 2.1. The
proof is simple. Suppose that y1, y2 are fundamental solutions of the second order equation,
then the vector-valued functions
x (1)
y1
=
y ′
, x
1
(2)
y2
=
y ′
,
2
are solutions of the first order reduction system. Then holds,
w = det x (1) , x (2)
=
y1 y2
y ′ 1 y ′
= Wy1y2 .
2
Example 5.4.6: Compute the Wronskian of the vector-valued functions given in Example
5.4.5, that is, x (1)
1
= e
1
−2t and x (2)
−1
= e
1
4t .
Solution: The Wronskian is the determinant of the solution matrix, with the vectors
placed in any order. For example, we can choose the order x (1) , x (2) . If we choose the
order x (2) , x (1) , this second Wronskian is the negative of the first one. Choosing the first
order we get,
w(t) = det x (1) , x (2)
=
.
e−2t −e4t e−2t e4t = e−2t e 4t + e −2t e 4t .
We conclude that w(t) = 2e 2t . ⊳
We saw in Section 5.3, Theorem 5.3.12, that the determinant of an n × n matrix is non-zero
iff the matrix column vectors form a linearly independent set. Since in Example 5.4.6 the
Wronskian is non-zero for all t ∈ R, this implies that the solution set {x (1) , x (2) } is linearly
independent for every t ∈ R.
,

G. NAGY – ODE July 2, 2013 177
Example 5.4.7: Show that the vector-valued functions x (1)
3t e
=
2e3t
and x (2)
−t e
=
−2e−t
form a linearly independent set for all t ∈ R.
3t e e
Solution: We compute the determinant of the matrix X(t) =
−t
, that is,
w(t) =
e3t e−t 2e3t −2e−t 2e 3t −2e −t
= −2e2t − 2e 2t ⇒ w(t) = −4e 2t = 0 t ∈ R.
Wronskians are used to find linearly independent set of n solutions to n × n linear homogeneous
differential systems. These linearly independent solution sets contain very important
information about the system. We will see that they actually contain all the information
needed to construct all possible solutions to that system. Because of that we give such
solutions a name.
Definition 5.4.3. Consider the n × n linear homogeneous differential system x ′ = A(t) x,
where A is an n × n matrix-valued function. The set {x (1) , · · · , x (n) } of n solutions to this
differential system is called a fundamental set in the interval I ⊂ R iff the set is linearly
independent for all t ∈ I. The n × n matrix-valued function
X = x (1) , · · · , x (n)
is called a fundamental matrix of the system.
Using the notation above, the Wronskian of a fundamental set is given by w(t) = det X(t) .
Since a fundamental set is linearly independent, that means the Wronskian of the fundamental
matrix is non-zero for every t where the solutions are defined.
Example 5.4.8: Find two fundamental matrices for the linear homogeneous system in Example
5.4.5.
Solution: One fundamental matrix is simple to find, is it constructed with the solutions
given in Example 5.4.5, that is,
X = x (1) , x (2) ⇒
−2t e
X(t) =
4t −e
.
e −2t e 4t
A second fundamental matrix can be obtained multiplying by any non-zero constant each
solution above. For example, another fundamental matrix is
˜X = 2x (1) , 3x (2) ⇒ ˜
−2t 2e
X(t) =
2e
4t −3e −2t 3e4t
.
⊳
5.4.3. Abel’s Theorem for systems. The following result is a generalization of Abel’s
Theorem 2.1.9 in Section 2.1 to n solutions of an n×n linear homogeneous differential system.
This result shows an important property of the Wronskian: The solution Wronskian vanishes
at a single point iff it vanishes identically for all t ∈ R.
Theorem 5.4.4 (Abel for systems). Let A be an n×n continuous matrix-valued function,
let tr A be the trace of A, let {x (1) , · · · , x (n) } be a set of n solutions to the differential
system x ′ = A(t)x, and let t0 ∈ R be an arbitrary number. Then, the Wronskian function
w = det x (1) , · · · , x (n) is given by the expression
w(t) = w(t0) e α(t) , α(t) =
t
t0
tr A(τ) dτ.
⊳

178 G. NAGY – ODE july 2, 2013
In the case of a constant matrix A, the equation above for the Wronskian reduces to
w(t) = w(t0) e (tr A) (t−t0) .
Once again, this result implies that the solution Wronskian vanishes at a single point iff
it vanishes identically for all t ∈ R. One consequence is this: If n solutions to a linear
homogeneous differential system are linearly independent at the initial time t0, then they
remain linearly independent for every time t ∈ R.
Proof of Theorem 5.4.4: The proof is based in an identity satisfied by the determinant
of certain matrix-valued functions. The proof of this identity is quite involved, so we do
not provide it here. The identity is the following: Every n × n, differentiable, invertible,
matrix-valued function Z, with values Z(t) for t ∈ R, satisfies the identity:
d
det(Z) = det(Z) tr
dt
−1 d
Z
dt Z
We use this identity with any fundamental matrix X = x (1) , · · · , x (n) of the linear homogeneous
differential system x ′ = Ax. Recalling that the Wronskian w(t) = det X(t) , the
identity above says,
w ′ (t) = w(t) tr X −1 (t) X ′ (t) .
We now compute the derivative of the fundamental matrix,
X ′ = x (1) ′ , · · · , x (n) ′ = Ax (1) , · · · , Ax (n) = AX,
where the equation on the far right comes from the definition of matrix multiplication.
Replacing this equation in the Wronskian equation we get
w ′ (t) = w(t) tr X −1 AX = w(t) tr X X −1 A = w(t) tr (A),
where in the second equation above we used a property of the trace of three matrices:
tr (ABC) = tr (CAB) = tr (BCA). Therefore, we have seen that the Wronskian satisfies
the equation
w ′ (t) = tr A(t) w(t). (5.4.3)
This is a linear differential equation of a single function w : R → R. We integrate it using
the integrating factor method from Section 1.2. The result is
w(t) = w(t0) e α(t) , α(t) =
t
t0
.
tr A(τ) dτ.
This establishes the Theorem.
Example 5.4.9: Show that the Wronskian of the fundamental matrix constructed with the
solutions given in Example 5.4.4 satisfies Eq. (5.4.3) above.
Solution: In Example 5.4.4 we have shown that the vector-valued functions x (1)
2
= e
1
2t
and x (2)
1
= e
2
−t are solutions to the system x ′
3 −2
= Ax, where A = . The matrix
2 −2
2t 2e e
X(t) =
−t
e 2t 2e −t
is a fundamental matrix of the system, since its Wronskian is non-zero,
w(t) =
2e2t e−t
= 4et − e t ⇒ w(t) = 3e t .
e 2t 2e −t

G. NAGY – ODE July 2, 2013 179
We need to compute the right-hand side and the left-hand side of Eq. (5.4.3) and verify that
they coincide. We start with the left-hand side,
w ′ (t) = 3e t = w(t).
The right-hand side is
tr (A) w(t) = (3 − 2) w(t) = w(t).
Therefore, we have shown that w ′ (t) = tr (A) w(t). ⊳
5.4.4. Existence of fundamental solutions. We now show that every n × n linear homogeneous
differential system has a set of n fundamental solutions. Abel’s Theorem above
together with Theorem 5.1.5 in Section 5.1 are the crucial statements needed to show this
existence result. We also need to recall the Definition 5.1.4 of an initial value problem,
which in the case of a linear non-homogeneous differential system is the following: Given an
n × n matrix-valued function A, a source vector-valued function g, an initial n-vector x0,
and an initial time t0, find the vector-valued function x solution of
x ′ = Ax + g, x(t0) = x0.
The following statement says that a fundamental set of solutions to a linear differential
system can be found solving n initial value problems. We use the notation e (i) to denote
the i-th column vector of the identity matrix I, that is, I = e (1) , · · · , e (n) .
Theorem 5.4.5 (Fundamental solutions). If the n × n matrix-valued function A is
continuous on an open interval I ⊂ R, then the differential equation x ′ = A(t) x has a
fundamental set given by {x (1) , · · · , x (n) }, where the vector-valued function x (i) : I → R n ,
for i = 1, · · · , n, is the solution of the initial value problem
x (i) ′ = A(t) x (i) , x (i) (t0) = e (i) . (5.4.4)
Proof of Theorem 5.4.5: Since linear homogeneous differential systems are particular
cases of the differential systems in Picard-Lindelöf Theorem 5.1.5, each initial value problem
in Eq. (5.4.4) has a unique solution x (i) , for i = 1, · · · , n. These solutions form a
fundamental set iff the set is linearly independent. We can check whether these solutions
form an independent set or not computing their Wronskian,
w(t) = det x (1) , · · · , x (n) .
Abel’s Theorem 5.4.4 says that the Wronskian satisfies
w(t) = e α(t) w(t0), α(t) =
t
t0
tr A(τ) dτ.
No matter what function α actually is, the function values e α(t) = 0. Hence, w(t) = 0 iff
w(t 0) = 0. However, the Wronskian at the initial value t 0 is simple to compute using the
initial conditions in Eq. (5.4.4),
w(t 0) = det e (1) , · · · , e (n) ⇒ w(t 0) = 1.
Since w(t0) = 0, so is w(t) for all t, showing that the set {x (1) , · · · , x (n) } is a fundamental
set. This establishes the Theorem.
We now know that the Definition 5.4.3 of fundamental set is indeed a useful one, since
every linear homogeneous differential system has at least one fundamental set. The next
result explains why this solution set is called fundamental. We now show that knowing a
fundamental set is equivalent to knowing all solutions to a system. The reason is that every
solution of a linear homogeneous differential system can be expressed as a particular linear
combination of the solutions in a given fundamental set.

180 G. NAGY – ODE july 2, 2013
Theorem 5.4.6. If A is a continuous n×n matrix-valued function, and the set of n-vectorvalued
functions {x (1) , · · · , x (n) } is a fundamental set of the n×n linear system x ′ = A(t) x,
then for every solution, x, to this differential system there exists a unique set of constants
{c1, · · · , cn} such that the following expression holds,
x(t) = c 1x (1) (t) + · · · + cnx (n) (t).
Proof of Theorem 5.4.6: We know that each function x, x (1) , · · · , x (n) is solution of the
differential equation x ′ = A(t) x. This means that for every t where the solutions are defined
we can construct the set {x, x (1) , · · · , x (n) } containing n+1 vectors in R n . From basic results
in Linear Algebra we know that this set must be linearly dependent. An example is a set of
three vectors on a plane; that set must be linearly dependent. Therefore, there exist n + 1
numbers, c 0(t), ˜c 1(t), · · · , ˜cn(t), which may depend on t, not all of them zero, such that,
c0(t) x(t) + ˜c1(t) x (1) (t) · · · + ˜cn(t) x (n) (t) = 0.
Since the set {x (1) (t), · · · , x (n) (t)} is linearly independent for all t, then the number c0(t)
must be non-zero for all t. Otherwise, all functions c 0, · · · , cn must vanish, which we just
said it is not the case. So, re-ordering terms and dividing by c 0(t) we get the equation
x(t) = c1(t) x (1) (t) · · · + cn(t) x (n) (t), (5.4.5)
where ci(t) = −˜ci(t)/c0(t), for i = 1, · · · , n. It can be shown that every solution of
x ′ = A(t) x must be differentiable, since A is continuous. Knowing that all functions
x, x (1) , · · · , x (n) are differentiable implies that the functions c1, · · · , cn must be differentiable
too. Now compute the derivative with respect to t of the equation above. We get,
x ′ = c1 x (1) ′ + · · · + cn x (n) ′ + c ′ 1 x (1) · · · + c ′ n x (n) .
Recalling that all the vector-valued functions above are solution to the differential equation,
we can replace their derivatives by multiplication by matrix A, that is,
A(t) x = c1 A(t) x (1) + · · · + cn A(t) x (n) + c ′ 1 x (1) · · · + c ′ n x (n) .
Using that the matrix-vector product is a linear operation, we get,
Recalling Eq. (5.4.5), we get
A(t) x = A(t) c1 x (1) + · · · + cn x (n) + c ′ 1 x (1) · · · + c ′ n x (n) .
A(t) x = A(t) x + c ′ 1 x (1) · · · + c ′ n x (n) ⇔ c ′ 1 x (1) · · · + c ′ n x (n) = 0.
Use again that the set {x (1) (t), · · · , x (n) (t)} is linearly independent for all t, we get that
c ′ 1 = 0, · · · c ′ n = 0.
So every function ci must be a constant. Then Eq. (5.4.5) now has the form
x(t) = c1 x (1) (t) · · · + cn x (n) (t),
with constants c1, · · · , cn. We only need to show that this set of constants is unique. Suppose
that we have a second set of constants ĉ1, · · · , ĉn such that
x(t) = ĉ1 x (1) (t) · · · + ĉn x (n) (t).
Subtract the latter expression from the former,
(c1 − ĉ1) x (1) (t) · · · + (cn − ĉn) x (n) (t) = 0.
One last time we use that the set of vectors {x (1) (t), · · · , x (n) (t)} is linearly independent for
all t, then we get that ci = ĉi. The constants are unique. This establishes the Theorem.
The Theorem above is the reason for the following definition.

G. NAGY – ODE July 2, 2013 181
Definition 5.4.7. The general solution of an n×n linear homogeneous differential system
x ′ = Ax is the set of n-vector-valued functions x(t) = c1 x (1) (t) · · · + cn x (n) (t), for every
constants c1, · · · , cn.
Example 5.4.10: Find the general solution to differential equation in Example 5.4.4 and
then use this general solution to find the solution of the initial value problem
x ′
1
3 −2
= Ax, x(0) = , A = .
5
2 −2
Solution: From Example 5.4.4 we know that the general solution of the differential equation
above can be written as
2
x(t) = c1 e
1
2t
1
+ c2 e
2
−t .
Before imposing the initial condition on this general solution, it is convenient to write this
general solution in terms of the fundamental matrix of the equation above, that is,
2t 2e e
x(t) =
−t
c1
⇔ x(t) = X(t)c,
e 2t 2e −t
where we introduced the fundamental and the constant vector
2t 2e e
X(t) =
−t
c1
, c = .
c2
e 2t 2e −t
The initial condition fixes the vector c, that is, its components c1, c2, as follows,
x(0) = X(0) c ⇒ c = X(0) −1 x(0).
Since the fundamental matrix at t = 0 has the form,
2 1
X(0) = ⇒
1 2
X(0) −1 1
=
3
introducing X(0) −1 in the equation for c above we get
c = 1
2
3 −1
−1 1 −1
=
2 5 3
⇒
c2
2 −1
,
−1 2
c1 = −1,
c 2 = 3.
We conclude that the solution to the initial value problem above is given by
2
x(t) = − e
1
2t
1
+ 3 e
2
−t .
⊳

182 G. NAGY – ODE july 2, 2013
5.4.5. Exercises.
5.4.1.- . 5.4.2.- .

G. NAGY – ODE July 2, 2013 183
5.5. Diagonalizable matrices
We continue with the review on Linear Algebra we started in Sections 5.2 and 5.3. We
have seen that a square matrix is a function on the space of vectors. The matrix acts on a
vector and the result is another vector. In this section we see that, given an n × n matrix,
there may exist lines in R n that are left invariant under the action of the matrix. The
matrix acting on a non-zero vectors in such line is proportional to that vector. The vector
is called an eigenvector of the matrix, and the proportionality factor is called an eigenvalue.
If an n × n matrix has a linearly independent set containing n eigenvectors, then we call
that matrix diagonalizable. In the next Section we will study linear differential systems
with constant coefficients having a diagonalizable coefficients matrix. We will see that it is
fairly simple to find solutions to such differential systems. The solutions can be expressed in
terms of the exponential of the coefficient matrix. For that reason we study in this Section
how to compute exponentials of square diagonalizable matrices.
5.5.1. Eigenvalues and eigenvectors. When a square matrix acts on a vector the result
is another vector that, more often than not, points in a different direction from the original
vector. However, there may exist vectors whose direction is not changed by the matrix. We
give these vectors a name.
Definition 5.5.1. A non-zero n-vector v and a number λ are respectively called an eigenvector
and eigenvalue of an n × n matrix A iff the following equation holds,
Av = λv.
We see that an eigenvector v determines a particular direction in the space R n , given by
(av) for a ∈ R, that remains invariant under the action of the function given by matrix A.
That is, the result of matrix A acting on any vector (av) on the line determined by v is
again a vector on the same line, as the following calculation shows it,
A(av) = a Av = aλv = λ(av).
Example 5.5.1: Verify that the pair λ1, v1 and the pair λ2, v2 are eigenvalue and eigenvector
pairs of matrix A given below,
A =
1 3
,
3 1
⎧
1
⎪⎨ λ1 = 4 v1 = ,
1
⎪⎩
−1
λ2 = −2 v2 = .
1
Solution: We just must verify the definition of eigenvalue and eigenvector given above.
We start with the first pair,
1 3 1 4 1
Av1 =
= = 4 = λ1v1
3 1 1 4 1
⇒ Av1 = λ1v1.
A similar calculation for the second pair implies,
1 3 −1 2 −1
Av2 =
= = −2 = λ
3 1 1 −2 1
2v2 ⇒ Av2 = λ2v2. Example 5.5.2: Find the eigenvalues and eigenvectors of the matrix A =
0 1
.
1 0
Solution: This is the matrix given in Example 5.3.1. The action of this matrix on the
plane is a reflection along the line x 1 = x 2, as it was shown in Fig. 24. Therefore, this line
⊳

184 G. NAGY – ODE july 2, 2013
x 1 = x 2 is left invariant under the action of this matrix. This property suggests that an
eigenvector is any vector on that line, for example
1 0 1 1 1
v1 = ⇒
= ⇒ λ1 = 1.
1 1 0 1 1
1
So, we have found one eigenvalue-eigenvector pair: λ1 = 1, with v1 = . We remark that
1
any non-zero vector proportional to v1 is also an eigenvector. Another choice fo eigenvalue-
3
eigenvector pair is λ1 = 1, with v1 = . It is not so easy to find a second eigenvector
3
which does not belong to the line determined by v1. One way to find such eigenvector is
noticing that the line perpendicular to the line x1 = x2 is also left invariant by matrix A.
Therefore, any non-zero vector on that line must be an eigenvector. For example the vector
v2 below, since
−1 0 1 −1 1 −1
v2 = ⇒
= = (−1) ⇒ λ2 = −1.
1 1 0 1 −1
1
−1
So, we have found a second eigenvalue-eigenvector pair: λ2 = −1, with v2 = . These
1
two eigenvectors are displayed on Fig. 26. ⊳
v
2
x
2
Ax
x = x
Av = v
1
1 2
2
1
Av = −v
2 2
x
x 1
Figure 26. The first picture shows the eigenvalues and eigenvectors of
the matrix in Example 5.5.2. The second picture shows that the matrix
in Example 5.5.3 makes a counterclockwise rotation by an angle θ, which
proves that this matrix does not have eigenvalues or eigenvectors.
There exist matrices that do not have eigenvalues and eigenvectors, as it is show in the
example below.
cos(θ) − sin(θ)
Example 5.5.3: Fix any number θ ∈ (0, 2π) and define the matrix A =
.
sin(θ) cos(θ)
Show that A has no real eigenvalues.
Solution: One can compute the action of matrix A on several vectors and verify that the
action of this matrix on the plane is a rotation counterclockwise by and angle θ, as shown
in Fig. 26. A particular case of this matrix was shown in Example 5.3.2, where θ = π/2.
Since eigenvectors of a matrix determine directions which are left invariant by the action of
the matrix, and a rotation does not have such directions, we conclude that the matrix A
above does not have eigenvectors and so it does not have eigenvalues either. ⊳
x
Ax
0
x
x 1

G. NAGY – ODE July 2, 2013 185
Remark: We will show that matrix A in Example 5.5.3 has complex-valued eigenvalues.
We now describe a method to find eigenvalue-eigenvector pairs of a matrix, if they exit.
In other words, we are going to solve the eigenvalue-eigenvector problem: Given an n × n
matrix A find, if possible, all its eigenvalues and eigenvectors, that is, all pairs λ and v = 0
solutions of the equation
Av = λv.
This problem is more complicated than finding the solution x to a linear system Ax = b,
where A and b are known. In the eigenvalue-eigenvector problem above neither λ nor v are
known. To solve the eigenvalue-eigenvector problem for a matrix A we proceed as follows:
(a) First, find the eigenvalues λ;
(b) Second, for each eigenvalue λ find the corresponding eigenvectors v.
The following result summarizes a way to solve the steps above.
Theorem 5.5.2 (Eigenvalues-eigenvectors).
(a) The number λ is an eigenvalue of an n × n matrix A iff holds,
det(A − λI) = 0. (5.5.1)
(b) Given an eigenvalue λ of an n × n matrix A, the corresponding eigenvectors v are the
non-zero solutions to the homogeneous linear system
(A − λI)v = 0. (5.5.2)
Proof of Theorem 5.5.2: The number λ and the non-zero vector v are an eigenvalueeigenvector
pair of matrix A iff holds
Av = λv ⇔ (A − λI)v = 0,
where I is the n × n identity matrix. Since v = 0, the last equation above says that the
columns of the matrix (A − λI) are linearly dependent. This last property is equivalent, by
Theorem 5.3.12, to the equation
det(A − λI) = 0,
which is the equation that determines the eigenvalues λ. Once this equation is solved,
substitute each solution λ back into the original eigenvalue-eigenvector equation
(A − λI)v = 0.
Since λ is known, this is a linear homogeneous system for the eigenvector components. It
always has non-zero solutions, since λ is precisely the number that makes the coefficient
matrix (A − λI) not invertible. This establishes the Theorem.
1 3
Example 5.5.4: Find the eigenvalues λ and eigenvectors v of the matrix A = .
3 1
Solution: We first find the eigenvalues as the solutions of the Eq. (5.5.1). Compute
1 3 1 0 1 3 λ 0 (1 − λ) 3
A − λI = − λ = − =
.
3 1 0 1 3 1 0 λ 3 (1 − λ)
Then we compute its determinant,
0 = det(A − λI) = (1
− λ) 3
3 (1 − λ) = (λ − 1)2 − 9 ⇒
λ+ = 4,
λ- = −2.
We have obtained two eigenvalues, so now we introduce λ + = 4 into Eq. (5.5.2), that is,
1 − 4 3 −3 3
A − 4I =
=
.
3 1 − 4 3 −3

186 G. NAGY – ODE july 2, 2013
Then we solve for v + the equation
(A − 4I)v + = 0 ⇔
−3 3
3 −3
v +
1
v +
2
=
0
.
0
The solution can be found using Gauss elimination operations, as follows,
+
−3 3 1 −1 1 −1
v 1 = v
→ →
⇒
3 −3 3 −3 0 0
+
2,
free.
Al solutions to the equation above are then given by
v +
1
= =
1
v +
2
v +
2
v +
2 ⇒ v + =
1
,
1
where we have chosen v +
2 = 1. A similar calculation provides the eigenvector v - associated
with the eigenvalue λ- = −2, that is, first compute the matrix
3 3
A + 2I =
3 3
then we solve for v - the equation
(A + 2I)v - = 0 ⇔
3 3
3 3
v -
1
v -
2
=
v +
2
0
.
0
The solution can be found using Gauss elimination operations, as follows,
-
3 3 1 1 1 1
v1 = −v
→ → ⇒
3 3 3 3 0 0
-
2,
free.
Al solutions to the equation above are then given by
v -
−1
= =
1
−v -
2
v -
2
v -
2 ⇒ v - =
v -
2
−1
,
1
where we have chosen v -
2 = 1. We therefore conclude that the eigenvalues and eigenvectors
of the matrix A above are given by
λ+ = 4, v + =
1
, λ- = −2, v
1
- =
−1
.
1
It is useful to introduce few more concepts, that are common in the literature.
Definition 5.5.3. The characteristic polynomial of an n × n matrix A is the function
p(λ) = det(A − λI).
Example 5.5.5: Find the characteristic polynomial of matrix A =
1 3
.
3 1
Solution: We need to compute the determinant
p(λ) = det(A − λI) = (1
− λ) 3
3 (1 − λ) = (1 − λ)2 − 9 = λ 2 − 2λ + 1 − 9.
We conclude that the characteristic polynomial is p(λ) = λ 2 − 2λ − 8. ⊳
Since the matrix A in this example is 2 × 2, its characteristic polynomial has degree two.
One can show that the characteristic polynomial of an n × n matrix has degree n. The
eigenvalues of the matrix are the roots of the characteristic polynomial. Different matrices
may have different types of roots, so we try to classify these roots in the following definition.
⊳

G. NAGY – ODE July 2, 2013 187
Definition 5.5.4. Given an n × n matrix A with eigenvalues λi, with i = 1, · · · , k n, it
is always possible to express the matrix characteristic polynomial as
p(λ) = (λ − λ 1) r1 · · · (λ − λk) rk .
The number ri is called the algebraic multiplicity of the eigenvalue λi. Furthermore, the
geometric multiplicity of the eigenvalue λi, denoted as si, is the maximum number of
eigenvectors vectors corresponding to that eigenvalue λi forming a linearly independent set.
Example 5.5.6: Find the eigenvalues algebraic and geometric multiplicities of the matrix
1 3
A =
3 1
Solution: In order to find the algebraic multiplicity of the eigenvalues we need first to find
the eigenvalues. We now that the characteristic polynomial of this matrix is given by
p(λ) = (1
− λ) 3
3 (1 − λ) = (λ − 1)2 − 9.
The roots of this polynomial are λ1 = 4 and λ2 = −2, so we know that p(λ) can be rewritten
in the following way,
p(λ) = (λ − 4)(λ + 2).
We conclude that the algebraic multiplicity of the eigenvalues are both one, that is,
λ1 = 4, r1 = 1, and λ2 = −2, r2 = 1.
In order to find the geometric multiplicities of matrix eigenvalues we need first to find the
matrix eigenvectors. This part of the work was already done in the Example 5.5.4 above
and the result is
λ1 = 4, v (1) =
1
, λ2 = −2, v
1
(2) =
−1
.
1
From this expression we conclude that the geometric multiplicities for each eigenvalue are
just one, that is,
λ1 = 4, s 1 = 1, and λ2 = −2, s 2 = 1.
⊳
The following example shows that two matrices can have the same eigenvalues, and so the
same algebraic multiplicities, but different eigenvectors with different geometric multiplicities.
⎡
3 0
⎤
1
Example 5.5.7: Find the eigenvalues and eigenvectors of the matrix A = ⎣0
3 2⎦.
0 0 1
Solution: We start finding the eigenvalues, the roots of the characteristic polynomial
(3
− λ) 0 1
p(λ) =
0 (3 − λ) 2
0 0 (1 − λ) = −(λ − 1)(λ − 3)2
λ1 = 1, r1 = 1,
⇒
λ2 = 3, r2 = 2.
We now compute the eigenvector associated with the eigenvalue λ1 = 1, which is the solution
of the linear system
(A − I)v (1) = 0 ⇔
⎡
2
⎣0
0
2
⎤ ⎡
1 v
2⎦
⎢
⎣
0 0 0
(1)
1
v (1)
⎤ ⎡ ⎤
0
⎥
⎦ = ⎣0⎦
.
0
2
v (1)
3

188 G. NAGY – ODE july 2, 2013
After the few Gauss elimination operation we obtain the following,
⎡
2
⎣0 0
0
2
0
⎤ ⎡
1
2⎦
→ ⎣
0
1 0 0
0
1
0
⎤
1
2
1⎦
0
⇒
⎧
⎪⎨
v
⎪⎩
(1)
1 = − v(1) 3
2 ,
v (1)
2 = −v (1)
3 ,
free.
Therefore, choosing v (1)
3
v (1)
3
= 2 we obtain that
v (1) ⎡ ⎤
−1
= ⎣−2⎦
,
2
λ1 = 1, r1 = 1, s1 = 1.
In a similar way we now compute the eigenvectors for the eigenvalue λ2 = 3, which are all
solutions of the linear system
(A − 3I)v (2) = 0 ⇔
⎡
0
⎣0
0
0
⎤ ⎡
1 v
2 ⎦ ⎢
⎣
0 0 −2
(2)
1
v (2)
⎤ ⎡ ⎤
0
⎥
⎦ = ⎣0⎦
.
0
2
v (2)
3
After the few Gauss elimination operation we obtain the following,
⎡
0
⎣0
0
0
0
0
⎤ ⎡
1 0
2 ⎦ → ⎣0
−2 0
0
0
0
⎤
1
0⎦
0
⇒
⎧
⎪⎨
v
⎪⎩
(2)
1 free,
v (2)
2 free,
= 0.
Therefore, we obtain two linearly independent solutions, the first one v (2) with the choice
v (2)
1 = 1, v (2)
2 = 0, and the second one w (2) with the choice v (2)
1 = 0, v (2)
2 = 1, that is
v (2) ⎡
= ⎣ 1
⎤
0⎦
, w
0
(2) ⎡
= ⎣ 0
⎤
1⎦
, λ2 = 3, r2 = 2, s2 = 2.
0
Summarizing, the matrix in this example has three linearly independent eigenvectors. ⊳
⎡
3 1
⎤
1
Example 5.5.8: Find the eigenvalues and eigenvectors of the matrix A = ⎣0
3 2⎦.
0 0 1
Solution: Notice that this matrix has only the coefficient a12 different from the previous
example. Again, we start finding the eigenvalues, which are the roots of the characteristic
polynomial
(3
− λ) 1 1
p(λ) =
0 (3 − λ) 2
0 0 (1 − λ) = −(λ − 1)(λ − 3)2
λ1 = 1, r1 = 1,
⇒
λ2 = 3, r2 = 2.
So this matrix has the same eigenvalues and algebraic multiplicities as the matrix in the
previous example. We now compute the eigenvector associated with the eigenvalue λ1 = 1,
which is the solution of the linear system
⎡ ⎤ ⎡
2 1 1 v
⎢
(1)
⎤ ⎡ ⎤
1 0
⎥
(A − I)v (1) = 0 ⇔
⎣0
2 2⎦
⎣
0 0 0
v (1)
v (2)
3
2
v (1)
3
⎦ = ⎣0⎦
.
0

G. NAGY – ODE July 2, 2013 189
After the few Gauss elimination operation we obtain the following,
⎧
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
2 1 1 1 1 1 1 0 0 ⎪⎨
⎣0
2 2⎦
→ ⎣0
1 1⎦
→ ⎣0
1 1⎦
⇒
0 0 0 0 0 0 0 0 0 ⎪⎩
Therefore, choosing v (1)
3
v (1)
1
v (1)
2
v (1)
3
= 1 we obtain that
v (1) ⎡ ⎤
0
= ⎣−1⎦
,
1
λ1 = 1, r1 = 1, s1 = 1.
= 0,
= −v (1)
3 ,
In a similar way we now compute the eigenvectors for the eigenvalue λ2 = 3. However, in
this case we obtain only one solution, as this calculation shows,
(A − 3I)v (2) ⎡ ⎤ ⎡
0 1 1 v
= 0 ⇔ ⎣0
0 2 ⎦ ⎢
⎣
0 0 −2
(2)
1
v (2)
⎤ ⎡ ⎤
0
⎥
⎦ = ⎣0⎦
.
0
2
v (2)
3
After the few Gauss elimination operation we obtain the following,
⎧
⎡ ⎤ ⎡ ⎤
0 1 1 0 1 0 ⎪⎨
v
⎣0
0 2 ⎦ → ⎣0
0 1⎦
⇒
0 0 −2 0 0 0 ⎪⎩
(2)
1 free,
v (2)
2 = 0,
= 0.
Therefore, we obtain only one linearly independent solution, which corresponds to the choice
v (2)
1 = 1, that is,
v (2) ⎡
= ⎣ 1
⎤
0⎦
,
0
λ2 = 3, r2 = 2, s2 = 1.
Summarizing, the matrix in this example has only two linearly independent eigenvectors,
and in the case of the eigenvalue λ2 = 3 we have the strict inequality
1 = s2 < r2 = 2.
5.5.2. Diagonalizable matrices. We first introduce the notion of a diagonal matrix. Later
on we define the idea of a diagonalizable matrix as a matrix that can be transformed into a
diagonal matrix by a simple transformation.
⎡
⎤
Definition 5.5.5. An n × n matrix A is called diagonal iff holds A =
v (2)
3
free.
⎢
⎣
⊳
a11
.
· · ·
. ..
0
.
⎥
⎦.
0 · · · ann
That is, a matrix is diagonal iff every non-diagonal coefficient vanishes. From now on we
use the following notation for a diagonal matrix A:
A = diag ⎡
a11
⎢
a11, · · · , ann = ⎣ .
· · ·
. ..
0
.
⎤
⎥
⎦ .
0 · · · ann
This notation says that the matrix is diagonal and shows only the diagonal coefficients,
since any other coefficient vanishes. Diagonal matrices are simple to manipulate since they
share many properties with numbers. For example the product of two diagonal matrices is

190 G. NAGY – ODE july 2, 2013
commutative. It is simple to compute power functions of a diagonal matrix. It is also simple
to compute more involved functions of a diagonal matrix, like the exponential function.
Example 5.5.9: For every positive integer n find An
2 0
, where A = .
0 3
Solution: We start computing A2 as follows,
A 2
2 0 2 0
= AA =
=
0 3 0 3
2 2 0
0 32
.
We now compute A3 ,
A 3 = A 2
2 2 0
A =
0 32
3 2 0 2 0
=
0 3 0 33
.
Using induction, it is simple to see that An
n 2 0
=
0 3n
. ⊳
Many properties of diagonal matrices are shared by diagonalizable matrices. These are
matrices that can be transformed into a diagonal matrix by a simple transformation.
Definition 5.5.6. An n × n matrix A is called diagonalizable iff there exists an invertible
matrix P and a diagonal matrix D such that
A = P DP −1 .
Systems of linear differential equations are simple to solve in the case that the coefficient
matrix is diagonalizable. In such case, it is simple to decouple the differential equations.
One solves the decoupled equations, and then transforms back to the original unknowns.
Example 5.5.10: We will see later on that matrix A is diagonalizable while B is not, where
1 3
A = , B =
3 1
1
3 1
.
2 −1 5
⊳
There is a deep relation between the eigenvalues and eigenvectors of a matrix and the
property of diagonalizability.
Theorem 5.5.7 (Diagonalizable matrices). An n × n matrix A is diagonalizable iff
matrix A has a linearly independent set of n eigenvectors. Furthermore,
A = P DP −1 , P = [v1, · · · , vn], D = diag
λ1, · · · , λn ,
where λi, vi, for i = 1, · · · , n, are eigenvalue-eigenvector pairs of matrix A.
Proof of Theorem 5.5.7:
(⇒) Since matrix A is diagonalizable, there exist an invertible matrix P and a diagonal
matrix D such that A = P DP −1 . Multiply this equation by P −1 on the left and by P on
the right, we get
D = P −1 AP. (5.5.3)
Since n × n matrix D is diagonal, it has a linearly independent set of n eigenvectors, given
by the column vectors of the identity matrix, that is,
De (i) = diie (i) , D = diag
d11, · · · , dnn , I = (1) (n) e , · · · , e .
So, the pair dii, e (i) is an eigenvalue-eigenvector pair of D, for i = 1 · · · , n. Using this
information in Eq. (5.5.3) we get
(i)
P e ,
diie (i) = De (i) = P −1 AP e (i) ⇒ A P e (i) = dii

G. NAGY – ODE July 2, 2013 191
where the last equation comes from multiplying the former equation by P on the left. This
last equation says that the vectors v (i) = P e (i) are eigenvectors of A with eigenvalue dii.
By definition, v (i) is the i-th column of matrix P , that is,
P = v (1) , · · · , v (n) .
Since matrix P is invertible, the eigenvectors set {v (1) , · · · , v (n) } is linearly independent.
This establishes this part of the Theorem.
(⇐) Let λi, v (i) be eigenvalue-eigenvector pairs of matrix A, for i = 1, · · · , n. Now use the
eigenvectors to construct matrix P = v (1) , · · · , v (n) . This matrix is invertible, since the
eigenvector set {v (1) , · · · , v (n) } is linearly independent. We now show that matrix P −1 AP
is diagonal. We start computing the product
that is,
AP = A v (1) , · · · , v (n) = Av (1) , · · · , Av (n) , = λ 1v (1) · · · , λnv (n) .
P −1 AP = P −1 λ 1v (1) , · · · , λnv (n) = λ 1P −1 v (1) , · · · , λnP −1 v (n) .
At this point it is useful to recall that P −1 is the inverse of P ,
I = P −1 P ⇔ e (1) , · · · , e (n) = P −1 v (1) , · · · , v (n) = P −1 v (1) , · · · , P −1 v (n) .
We conclude that e (i) = P −1 v (i) , for i = 1 · · · , n. Using these equations in the equation for
P −1 AP we get
P −1 AP = λ1e (1) , · · · , λne (n) = diag
λ1, · · · , λn .
Denoting D = diag
−1 λ1, · · · , λn we conlude that P AP = D, or equivalently
.
A = P DP −1 , P = v (1) , · · · , v (n) , D = diag λ 1, · · · , λn
This means that A is diagonalizable. This establishes the Theorem.
1
Example 5.5.11: Show that matrix A =
3
3
is diagonalizable.
1
Solution: We known that the eigenvalue eigenvector pairs are
λ1 = 4,
1
v1 =
1
and λ2 = −2, v2 =
Introduce P and D as follows,
1 −1
P =
1 1
⇒ P −1 = 1
2
1 1
, D =
−1 1
We must show that A = P DP −1 . This is indeed the case, since
P DP −1
1
=
1
−1 4
1 0
0 1 1
−2 2 −1
1
.
1
−1
.
1
4 0
.
0 −2
P DP −1
4 2 1 1 1 2 1 1 1
=
=
4 −2 2 −1 1 2 −1 −1 1
We conclude, P DP −1
1 3
= ⇒ P DP
3 1
−1 = A, that is, A is diagonalizable. ⊳

192 G. NAGY – ODE july 2, 2013
Example 5.5.12: Show that matrix B = 1
2
3 1
is not diagonalizable.
−1 5
Solution: We first compute the matrix eigenvalues. The characteristic polynomial is
3
1
− λ
p(λ) = 2 2
− 1
3
5
5
= − λ − λ +
− λ
2 2
2 2 1
4 = λ2 − 4λ + 4.
The roots of the characteristic polynomial are computed in the usual way,
λ = 1
√
4 ± 16 − 16
2
⇒ λ = 2, r = 2.
We have obtained a single eigenvalue with algebraic multiplicity r = 2. The associated
eigenvectors are computed as the solutions to the equation (A − 2I)v = 0. Then,
⎡
3
⎢ − 2
(A − 2I) = ⎣ 2
−
1
2
1
2
⎡
⎤
⎢−
⎥ ⎢
5
⎦
= ⎢
− 2 ⎣
2 1
2
−
1
2
1
2
⎤
⎥
1
⎥
⎦
2
→
1
0
−1
0
⇒
1
v = ,
1
s = 1.
We conclude that the biggest linearly independent set of eigenvalues for the 2 × 2 matrix B
contains only one vector, insted of two. Therefore, matrix B is not diagonalizable. ⊳
Because of Theorem 5.5.7 it is important in applications to know whether an n×n matrix
has a linearly independent set of n eigenvectors. More often than not there is no simple way
to check this property other than to compute all the matrix eigenvectors. However, there is
a simple particular case: When the n × n matrix has n different eigenvalues. In such case
we do not need to compute the eigenvectors. The following result says that such matrix
always have a linearly independent set of n eigenvectors, and so, by Theorem 5.5.7, will be
diagonalizable.
Theorem 5.5.8 (Different eigenvalues). If an n × n matrix has n different eigenvalues,
then this matrix has a linearly independent set of n eigenvectors.
Proof of Theorem 5.5.8: Let λ 1, · · · , λn be the eigenvalues of an n × n matrix A,
all different from each other. Let v (1) , · · · , v (n) the corresponding eigenvectors, that is,
Av (i) = λiv (i) , with i = 1, · · · , n. We have to show that the set {v (1) , · · · , v (n) } is linearly
independent. We assume that the opposite is true and we obtain a contradiction. Let us
assume that the set above is linearly dependent, that is, there exists constants c1, · · · , cn,
not all zero, such that,
c1v (1) + · · · + cnv (n) = 0. (5.5.4)
Let us name the eigenvalues and eigenvectors such that c1 = 0. Now, multiply the equation
above by the matrix A, the result is,
c 1λ 1v (1) + · · · + cnλnv (n) = 0.
Multiply Eq. (5.5.4) by the eigenvalue λn, the result is,
c1λnv (1) + · · · + cnλnv (n) = 0.
Subtract the second from the first of the equations above, then the last term on the righthand
sides cancels out, and we obtain,
c1(λ1 − λn)v (1) + · · · + cn−1(λn−1 − λn)v (n−1) = 0. (5.5.5)

G. NAGY – ODE July 2, 2013 193
Repeat the whole procedure starting with Eq. (5.5.5), that is, multiply this later equation
by matrix A and also by λn−1, then subtract the second from the first, the result is,
c1(λ1 − λn)(λ1 − λn−1)v (1) + · · · + cn−2(λn−2 − λn)(λn−2 − λn−1)v (n−2) = 0.
Repeat the whole procedure a total of n − 1 times, in the last step we obtain the equation
c1(λ1 − λn)(λ1 − λn−1) · · · (λ1 − λ2)v (1) = 0.
Since all the eigenvalues are different, we conclude that c 1 = 0, however this contradicts our
assumption that c1 = 0. Therefore, the set of n eigenvectors must be linearly independent.
Example 5.5.13: Is matrix A =
1 1
diagonalizable?
1 1
Solution: We compute the matrix eigenvalues, starting with the characteristic polynomial,
p(λ) = (1
− λ) 1
1 (1 − λ) = (1 − λ)2 − 1 = λ 2 − 2λ ⇒ p(λ) = λ(λ − 2).
The roots of the characteristic polynomial are the matrix eigenvalues,
λ 1 = 0, λ 2 = 2.
The eigenvalues are different, so by Theorem 5.5.8, matrix A is diagonalizable. ⊳
5.5.3. The exponential of a matrix. Functions of a diagonalizable matrix are simple
to compute in the case that the function admits a Taylor series expansion. One of such
functions is the exponential function. In this Section we compute the exponential of a
diagonalizable matrix. However, we emphasize that this method can be trivially extended
to any function admitting a Taylor series expansion. A key result needed to compute the
Taylor series expansion of a matrix is the n-power of the matrix.
Theorem 5.5.9. If the n × n matrix A is diagonalizable, with invertible matrix P and
diagonal matrix D satisfying A = P DP −1 , then for every integer n 1 holds
A n = P D n P −1 . (5.5.6)
Proof of Theorem 5.5.9: It is not difficult to generalize the calculation done in Example
5.5.9 to obtain the n-th power of a diagonal matrix D = diag
d1, · · · , dn , and the result
is another diagonal matrix given by
.
D n = diag d n 1 , · · · , d n n
We use this result and induction in n to prove Eq.(5.5.6). Since the case n = 1 is trivially
true, we start computing the case n = 2. We get
A 2 = P DP −1 2 = P DP −1 P DP −1 = P DDP −1 ⇒ A 2 = P D 2 P −1 ,
that is, Eq. (5.5.6) holds for n = 2. Now, assuming that this Equation holds for n, we show
that it also holds for n + 1. Indeed,
A (n+1) = A n A = P D n P −1 P DP −1 = P D n P −1 P DP −1 = P D n DP −1 .
We conclude that A (n+1) = P D (n+1) P −1 . This establishes the Theorem.
The exponential function f(x) = e x is usually defined as the inverse of the natural
logarithm function g(x) = ln(x), which in turns is defined as the area under the graph of
the function h(x) = 1/x from 1 to x, that is,
ln(x) =
x
1
1
dy, x ∈ (0, ∞).
y

194 G. NAGY – ODE july 2, 2013
It is not clear how to extend to matrices this way of defining the exponential function on
real numbers. However, the exponential function on real numbers satisfies several identities
that can be used as definition for the exponential on matrices. One of these identities is the
Taylor series expansion
e x =
∞
k=0
x k
k!
= 1 + x + x2
2!
+ x3
3!
+ · · · .
This identity is the key to generalize the exponential function to diagonalizable matrices.
Definition 5.5.10. The exponential of a square matrix A is defined as the infinite sum
e A ∞ A
=
n
. (5.5.7)
n!
n=0
One must show that the definition makes sense, that is, that the infinite sum in Eq. (5.5.7)
converges. We show in these notes that this is the case when matrix A is diagonal and when
matrix A is diagonalizable. The case of non-diagonalizable matrix is more difficult to prove,
and we do not do it in these notes.
Theorem 5.5.11. If D = diag
D d1 dn
d1, · · · , dn , then holds e = diag e , · · · , e .
Proof of Theorem 5.5.11: It is simple to see that the infinite sum in Eq (5.5.7) converges
for diagonal matrices. Start computing
e D ∞ 1
∞ k 1
= diag d1, · · · , dn =
k!
k! diag (d1) k , · · · , (dn) k .
k=0
Since each matrix in the sum on the far right above is diagonal, then holds
e D ∞ (d1)
= diag
k ∞ (dn)
, · · · ,
k!
k
.
k!
k=0
∞ (di)
Now, each sum in the diagonal of matrix above satisfies
k=0
k
k! = edi . Therefore, we
arrive to the equation eD = diag ed1 , · · · , edn
. This establishes the Theorem.
Example 5.5.14: Compute eA
2
, where A =
0
0
.
7
Solution: Theorem 5.5.11 implies that eA
2 e
=
0
0
e7
. ⊳
The case of diagonalizable matrices is more involved and is summarized below.
Theorem 5.5.12. If the n × n matrix A is diagonalizable, with invertible matrix P and
diagonal matrix D satisfying A = P DP −1 , then the exponential of matrix A is given by
k=0
k=0
e A = P e D P −1 . (5.5.8)
The formula above says that to find the exponential of a diagonalizable matrix there is no
need to compute the infinite sum in Definition 5.5.10. To compute the exponential of a
diagonalizable matrix it is only needed to compute the product of three matrices. It also
says that to compute the exponential of a diagonalizable matrix we need to compute the
eigenvalues and eigenvectors of the matrix.

k=0
G. NAGY – ODE July 2, 2013 195
Proof of Theorem 5.5.12: We again start with Eq. (5.5.7),
e A ∞ 1
=
k! An ∞ 1
=
k! (P DP −1 ) n ∞ 1
=
k! (P DnP −1 ),
k=0
where the last step comes from Theorem 5.5.9. Now, in the expression on the far right we
can take common factor P on the left and P −1 on the right, that is,
e A ∞ 1
= P
k! Dn P −1 .
k=0
The terms between brackets sum up to the exponential of the diagonal matrix D, that is,
e A = P e D P −1 .
Since the exponential of a diagonal matrix is computed in Theorem 5.5.11, this establishes
the Theorem.
Example 5.5.15: Compute e
A
1
, where A =
3
3
.
1
Solution: To compute the exponential of matrix A above we need the decomposition
A = P DP −1 . Matrices P and D are constructed with the eigenvectors and eigenvalues of
k=0
matrix A, respectively. From Example 5.5.4 we know that
1
λ1 = 4, v1 =
1
Introduce P and D as follows,
1 −1
P =
1 1
⇒ P −1 = 1
2
Then, the exponential function is given by
e At = P e Dt P −1 =
and λ2 = −2, v2 =
1 −1
1 1
1 1
, D =
−1 1
e 4t 0
0 e −2t
1
2
−1
.
1
4 0
.
0 −2
1 1
.
−1 1
Usually one leaves the function in this form. If we multiply the three matrices out we get
e At = 1
4t −2t 4t −2t (e + e ) (e − e )
2 (e4t − e−2t ) (e4t + e−2t
.
)
⊳
The exponential of an n × n matrix A can be used to define the matrix-valued function
with values e At . In the case that the matrix A is diagonalizable we obtain
e At = P e Dt P −1 ,
where eDt = diag ed1t , · · · , ednt and D = diag
d1, · · · , dn . It is not difficult to show that
de At
dt = A eAt = e At A.
This exponential function will be useful to express solutions of a linear homogeneous differential
system x ′ = Ax. This is the subject of the next Section.

196 G. NAGY – ODE july 2, 2013
5.5.4. Exercises.
5.5.1.- . 5.5.2.- .

G. NAGY – ODE July 2, 2013 197
5.6. Constant coefficient differential systems
In the rest of this Chapter we study linear differential systems with constant coefficients.
More precisely, given an n × n constant matrix A, an n-vector valued function g, and a
constant n-vector x0, find an n-vector valued function x solution to the initial value problem
x ′ (t) = A x(t) + g(t), x(t 0) = x 0. (5.6.1)
We have mentioned in Section 5.1 that the Picard-Lindelöf Theorem 5.1.5 applies to linear
differential systems, hence the initial value problem above always has a unique solution.
However, we have not said in a precise way how to obtain such solution. In this Section we
present a formula for the solution of the initial value problem in Eq. (5.6.1) in the case that
the coefficient matrix A is diagonalizable. We start studying the particular case of homogeneous
equations, that is, g = 0. We then study non-homogeneous equations with constant
source vector g. We finally study non-homogeneous equations with continuous source function
g. The expression for the solutions in these cases will be a generalization of the solution
for linear, constant coefficients, scalar equations given in Section 1.1, Eq. (1.1.10).
5.6.1. Diagonalizable homogeneous systems. We start studying the particular case of
homogeneous linear differential systems. We restrict our study to systems with diagonalizable
coefficient matrices. In this case it is easy to show that the solutions to differential
systems are given by a simple generalization of the solution formula for scalar differential
equations, Eq. (1.1.10) in the case b = 0. The solution formula for differential systems involves
the exponential of the equation coefficients, that is, the exponential of the coefficient
matrix.
Theorem 5.6.1 (Exponential expression). If the n × n constant matrix A is diagonalizable,
then the homogeneous initial value problem
x ′ (t) = A x(t), x(t0) = x0, (5.6.2)
has a unique solution for every initial condition x0 ∈ R n given by
x(t) = e A(t−t0) x0. (5.6.3)
Because the coefficient matrix A is diagonalizable, we can transform the coupled system
for the unknown vector components xi into a decoupled system for appropriately chosen
unknonwns. We solve each of the n decoupled equations using Eq. (1.1.10), and then we
transform this solution back to the original unknown vector components xi.
Proof of Theorem 5.6.1: Since the coefficient matrix A is diagonalizable, there exist an
n×n invertible matrix P and an n×n diagonal matrix D such that A = P DP −1 . Introducing
this information into the differential equation in Eq. (1.1.10), and then multiplying the whole
equation by P −1 , we obtain,
P −1 x ′ (t) = P −1 P DP −1 x(t).
Since matrix A is constant, so is P and D. In particular P −1 x ′ = P −1 x ′ . Therefore,
(P −1 x) ′ = D P −1 x .
Introduce the new unknown function y = P −1 x , then the equation above has the form
y ′ (t) = D y(t).
Now transform the initial condition too, that is, P −1 x(t 0) = P −1 x 0. Introducing the notation
y 0 = P −1 x0, we get the initial condition
y(t 0) = y 0.

198 G. NAGY – ODE july 2, 2013
Since the coefficient matrix D is diagonal, the initial value problem above for the unknown
y is decoupled. That is, expressing D = diag[λ1, · · · , λn], y 0 = [y0i], and y(t) = [yi(t)], then
for each i = 1, · · · , n holds that the initial value problem above has the form
y ′ i(t) = λi yi(t), yi(t 0) = y 0i.
Since the i-th equation involves only the unknown yi, we only need to solve n decoupled
scalar equations, each one for a single unknown function. The solution of each initial value
problem above can be found using Eq. (1.1.10), that is,
yi(t) = e λi(t−t0) y0i.
Once the equations are solved, we rewrite the solutions above back in vector notation,
y(t) = e D(t−t0) y0,
where we have used the expression e D(t−t0) = diag e λ1(t−t0) , · · · , e λn(t−t0) . We now multiply
the whole equation by matrix P and we recall that P −1 P = In, then
P y(t) = P e D(t−t0) (P −1 P )y0.
Recalling that P y = x, P y 0 = x0, and e At = P e Dt P −1 , we obtain
x(t) = e A(t−t0) x0.
This establishes the Theorem.
The expression for the solution given in Eq. (5.6.3) is beautiful, simple. However, the
exponential of a matrix requires some work to be computed, as the following example shows.
Example 5.6.1: Find the vector-valued function x solution to the differential system
x ′
3 1 2
= A x, x(0) = , A = .
2 2 1
Solution: First we need to find out whether the coefficient matrix A is diagonalizable or
not. Theorem 5.5.7 says that a 2 × 2 matrix is diagonalizable iff there exists a linearly
independent set of two eigenvectors. So we start computing the matrix eigenvalues, which
are the roots of the characteristic polynomial
p(λ) = det(A − λI2) = (1
− λ) 2
2 (1 − λ) = (1 − λ)2 − 4.
The roots of the characteristic polynomial are
(λ − 1) 2 = 4 ⇔ λ± = 1 ± 2 ⇔ λ+ = 3, λ− = −1.
The eigenvectors corresponding to the eigenvalue λ+ = 3 are the solutions v of the linear
system (A − 3I2)v (+) = 0. To find them, we perform Gauss operations on the matrix
−2 2 1 −1
A − 3I2 =
→
⇒ v1 = v2 ⇒ v
2 −2 0 0
(+)
1
= .
1
The eigenvectors corresponding to the eigenvalue λ− = −1 are the solutions v of the linear
system (A + I2)v (-) = 0. To find them, we perform Gauss operations on the matrix
2 2 1 1
A + I2 = → ⇒ v1 = −v2 ⇒ v
2 2 0 0
(-)
−1
= .
1
With this information and Theorem 5.5.7 we obtain that
A = P DP −1 , P =
1 −1
1 1
, D =
3 0
,
0 −1

and also that
so we conclude that
e At = P e Dt P −1 =
e At = 1
2
G. NAGY – ODE July 2, 2013 199
1
3t −1 e 0
1 1 0 e−t
1
2
3t −t 3t −t (e + e ) (e − e )
(e3t − e−t ) (e3t + e−t
.
)
The solution to the initial value problem above is,
x(t) = e At x0 = 1
3t −t 3t −t (e + e ) (e − e )
2 (e3t − e−t ) (e3t + e−t
3
) 2
1 1
,
−1 1
⇒ x(t) = 1
2
5e 3t + e −t
5e 3t − e −t
We now present an equivalent way to write down the solution function x given in Theorem
5.6.1, Eq. (5.6.3). This new expression for x looks simpler because it does not contain
the exponential of the system coefficient matrix. Only the matrix eigenvalues and a simple
linear combination of the matrix eigenvectors appear in this new expression.
Theorem 5.6.2 (Eigenpairs expression). If the n×n constant matrix A is diagonalizable,
with eigenpairs λi, v (i) for i = 1, · · · , n, then, the homogeneous initial value problem
has a unique for every initial condition x0 ∈ R n given by
.
x ′ (t) = A x(t) x(t 0) = x 0. (5.6.4)
x(t) = c1v (1) e λ1(t−t0) + · · · + cnv (n) e λn(t−t0) , (5.6.5)
where the constants c 1, · · · , cn are solution of the algebraic linear system
x0 = c1v (1) + · · · + cnv (n) . (5.6.6)
Just by looking at Eqs. (5.6.5)-(5.6.6) one can understand the reason for the simplicity of
this new expression for x; it is simple because it is not complete. The coefficients c1, · · · , cn
are said to be solutions of Eq. (5.6.6), but such solution is actually not computed. If one
solves the equations for c 1, · · · , cn, the resulting expression for x is the one in Eq. (5.6.3).
We think that Eq. (5.6.5) deserves to be highlighted in a Theorem, because is very common
in the literature, and simple to work with.
Proof of Theorem 5.6.2: Recall Theorem 5.5.7, where we showed that matrix A is
diagonalizable iff it can be written as
A = P DP −1 , P = v (1) , · · · , v (n) , D = diag[λ1, · · · , λn],
where λi, v (i) are eigenpairs of the matrix A. Let us use this information and the exponential
formula e At = P e Dt P −1 to rewrite the solution found in Eq. (5.6.3), that is,
x(t) = e At x0 = P e Dt P −1 x0.
Introduce the constant vector c = P −1 x0 and recall that
P e Dt = v (1) , · · · , v (n) diag e λ1t , · · · , e λnt = e λ1t v (1) , · · · , e λnt v (n) .
We then arrive at the expression
x(t) = e λ1t v (1) , · · · , e λnt v (n) c ⇔ x(t) = c 1e λ1t v (1) + · · · + cne λnt v (n) ,
where the constant vector c satisfies the equation P c = x0, that is,
c1v (1) + · · · + cnv (n) = x0.
This establishes the Theorem.
⊳

200 G. NAGY – ODE july 2, 2013
Remark: An alternative proof for Theorem 5.6.2 is simply to verify that x in Eq. (5.6.5) is
solution of the initial value problem in Eq. (5.6.4). Indeed, x is solution of the differential
equation, since for all constants c1, · · · , cn holds
Ax(t) = c 1Av (1) e λ1t + · · · + cnAv (n) e λnt
= c 1λ 1v (1) e λ1t + · · · + cnλnv (n) e λnt
= c1v (1) e λ1t ′ + · · · + cnv (n) e λnt ′
= x ′ (t).
The function x in Eq. (5.6.5) satisfies the initial condition because
x0 = x(0) = c1v (1) + · · · + cnv (n) ,
which is Eq. (5.6.6) for the constants c 1, · · · , cn.
Example 5.6.2: Find the vector-valued function x solution to the differential system
x ′
3 1 2
= A x, x(0) = , A = .
2 2 1
Solution: In Example 5.6.1 we have found the eigenvalues and eigenvectors of the coefficient
matrix, and the result is
λ1 = 3, v (1)
1
= ,
1
and λ2 = −1, v (2)
−1
= .
1
So, Eq. (5.6.5) implies that
x(t) = c 1
1
e
1
3t + c2
−1
e
1
−t ,
where the coefficients c1 and c2 are solutions of the initial condition equation
c1
c1
c1
1
1
+ c2
−1
1
=
3
2
⇒
1 −1
1 1
c2
=
3
2
⇒
c2
= 1
2
1 1
−1 1
3
.
2
We conclude that c1 = 5/2 and c2 = −1/2, so we reobtain the result in Example 5.6.1,
3t −t 5e + e
x(t) = 5
2
1
1
e 3t − 1
2
−1
1
e −t ⇔ x(t) = 1
2
5e 3t − e −t
5.6.2. The Fundamental Solutions. The expression in Eq. (5.6.5) for the solution of a
linear, homogeneous differential system says that there exist n different solutions to the
linear system x ′ = Ax. Just choose one of the coefficients ci non-zero and all the rest zero.
It is then useful to introduce the following definitions.
Definition 5.6.3. The set of functions x (1) , · · · , x (n) is called a fundamental set of
solutions to the homogeneous system x ′ = Ax iff each function x (i) for i = 1, · · · , n is solution
of the system above and the set is linealry independent for all t ∈ R. Each function x (i)
in a fundamental set is called a fundamental solution to the system above. Furthermore,
the matrix-valued function X = x (1) , · · · , x (n) is called the fundamental matrix of the
system above. Finally, the general solution of the differential equation above is the set of
all functions of the form
⎡ ⎤
x(t) = c1x (1) (t) + · · · + cnx (n) (t) ⇔ x(t) = X(t) c, c =
.
⎢
⎣
c1
.
cn
⎥
⎦ .
⊳

G. NAGY – ODE July 2, 2013 201
Any linearly independent set of n solutions to an n × n linear homogeneous differential
system defines a fundamental set. Hence, fundamental sets of solutions are not unique at all.
If the system coefficient matrix is diagonalizable, then Theorem 5.6.5 provides an explicit
expression for a fundamental set, a fundamental matrix and a general solution. They are
constructed with the eigenvalues and eigenvectors of the coefficient matrix.
Theorem 5.6.4. If the n × n constant matrix A is diagonalizable, with eigenpairs λi, v (i)
for i = 1, · · · , n, then, the linear homogeneous system x ′ = Ax has a fundamental set of
solutions given by
(1) (1) λ1t (n) (n) λnt
x (t) = v e , · · · , x (t) = v e
(5.6.7)
Then, the associated fundamental matrix is
X(t) = v (1) e λ1t , · · · , v (n) e λnt ,
and the general solution is given by every linear combination
x(t) = c1v (1) e λ1t + · · · + cnv (n) e λnt .
Remark: It is simple to see the relation between the fundamental solution matrix X of an
homogeneous system x ′ = Ax and the exponential matrix eAt . Indeed, if A = P DP −1 , then
X(t) = P eDt ; equivalently, eAt = X(t)P −1 . Also notice that X(0) = P .
Proof of Theorem 5.6.4: We have already seen in the Remark after Theorem 5.6.2 that
every function in the set given in (5.6.7) is solution to the system x ′ = Ax. We only need
to show that this set is linearly independent. For that reason we compute the determinant
of the matrix X. Since det P −1 = 0, we can do the following,
det X(t) = det X(t) detP −1
det P −1 = detX(t)P −1
det P −1
.
Recalling that X(t)P −1 = P e Dt P −1 we obtain that
det X(t) = det P e Dt P −1
det P −1
= det P det e Dt det P −1
det P −1
= det e Dt
det P −1.
Therefore, det X(t) = 0 iff det e Dt = 0. But this latter equatio is true, since
det e Dt = e λ1t · · · e λnt ⇔ det e Dt = e (λ1+···λn)t = 0.
Therefore det X(t) = 0 for all t ∈ R. This establishes the Theorem.
Remark: The function W (t) = det X(t) is called the Wronskian of a set of n solutions
to a differential system.
Example 5.6.3: Find the general solution to the 2 × 2 differential system
x ′
1 3
= Ax, A = .
3 1
Solution: Following Theorem 5.6.5, in order to find the solution to the equation above we
must start finding the eigenvalues and eigenvectors of the coefficient matrix A. This part
of the work was already done in Example 5.5.4. We have found that A has two linearly
independent eigenvectors, more precisely,
λ1 = 4, v (1)
1
=
⇒ x (1)
1
= e 4t ,
λ 2 = −2, v (2) =
1
−1
1
1
⇒ x (2)
−1
= e
1
−2t .

202 G. NAGY – ODE july 2, 2013
Therefore, the general solution of the differential equation is
1
x(t) = c1 e
1
4t
−1
+ c2 e
1
−2t , c1, c2 ∈ R.
Remark: We can write the general solution using the fundamental matrix X(t) and the
constant vector c,
4t e
X(t) =
−2t −e
,
c1
c = ,
e 4t e −2t
and the general solution is x(t) = X(t)c, that is,
x1(t)
=
x2(t) e 4t −e −2t
e 4t e −2t
c2
c1
.
c2 Example 5.6.4: Use the fundamental solution matrix found in Example 5.6.3 to find the
solution to the initial value problem
x ′
2
1 3
= Ax, x(0) = , A = .
4
3 1
Solution: We known from Example 5.6.3 that the general solution to the differential
equation above is given by
1
x(t) = c1 e
1
4t
−1
+ c2 e
1
−2t , c1, c2 ∈ R.
Equivalently, introducing the fundamental matrix X and the vector c as in Example 5.6.3,
4t e
X(t) =
e
−2t −e 4t e−2t
,
c1
c =
c2 ⇒ x(t) = X(t)c.
The initial condition is an equation for the constant vector c,
X(0)c = x(0) ⇔
1
1
−1 c1
=
1 c2
2
.
4
The solution of the linear system above can be expressed in terms of the inverse matrix
−1 1 1 1
X(0) = ,
2 −1 1
as follows,
c = X(0) −1 x(0) ⇔
c1
c 2
= 1
2
1 1
−1 1
2
4
So, the solution to the initial value problem is given by
1
x(t) = 3 e
1
4t
−1
+ e
1
−2t .
⇒
c1
=
c2
3
.
1
Remark: In the case that the coefficient matrix A of a linear homogeneous differential
system has n distinct eigenvalues, then Theorem 5.5.8 implies that A is diagonalizable, and
every the result in this Section applies to this case.
⊳
⊳

G. NAGY – ODE July 2, 2013 203
5.6.3. Diagonalizable systems with continuous sources. We continue our study of
diagonalizable linear differential systems with the case of non-homogeneous, continuous
sources. The solution of an initial value problem in this case involves again the exponential
of the coefficient matrix. The solution for these systems is a generalization of the solution
formula for single scalar equations, which is given by Eq. (1.2.7), in the case that the
coefficient function a is constant.
Theorem 5.6.5 (Exponential expression). If the n × n constant matrix A is diagonalizable
and the n-vector-valued function g is constant, then the initial value problem
x ′ (t) = A x(t) + g, x(t0) = x0.
has a unique solution for every initial condition x0 ∈ R n given by
A(t−t0)
x(t) = e x0 +
t
t0
e −A(τ−t0) g(τ) dτ
. (5.6.8)
Remark: In the case of an homogeneous system, g = 0, and we reobtain Eq. (5.6.3). In
the case that the coefficient matrix A is invertible and the source function g is constant, the
integral in Eq. (5.6.8) can be computed explicitly and the result is
x(t) = e A(t−t0) x0 − A −1 g − A −1 g.
The expression above is the generalizations for systems of Eq. (1.1.10) for scalar equations.
Proof of Theorem 5.6.5: Since the coefficient matrix A is diagonalizable, there exist an
n×n invertible matrix P and an n×n diagonal matrix D such that A = P DP −1 . Introducing
this information into the differential equation in Eq. (1.1.10), and then multiplying the whole
equation by P −1 , we obtain,
P −1 x ′ (t) = P −1 P DP −1 x(t) + P −1 g(t).
Since matrix A is constant, so is P and D. In particular P −1 x ′ = P −1 x ′ . Therefore,
(P −1 x) ′ = D P −1 x + P −1 g .
Introduce the new unknown function y = P −1 x and the new source function h = P −1 g ,
then the equation above has the form
y ′ (t) = D y(t) + h(t).
Now transform the initial condition too, that is, P −1 x(t 0) = P −1 x 0. Introducing the notation
y 0 = P −1 x 0, we get the initial condition
y(t0) = y 0.
Since the coefficient matrix D is diagonal, the initial value problem above for the unknown
y is decoupled. That is, expressing D = diag[λ1, · · · , λn], h = [hi], y 0 = [y0i], and y = [yi],
then for i = 1, · · · , n holds that the initial value problem above has the form
y ′ i(t) = λi yi(t) + hi(t), yi(t0) = y0i.
The solution of each initial value problem above can be found using Eq. (1.2.7), that is,
λi(t−t0)
yi(t) = e y0i +
t
t0
t0
e −λi(τ−t0) hi(τ) dτ
We rewrite the equations above in vector notation as follows,
t
D(t−t0)
y(t) = e y0 + e −D(τ−t0)
h(τ) dτ ,
.

204 G. NAGY – ODE july 2, 2013
where we recall that eD(t−t0)
λ1(t−t0) λn(t−t0) = diag e , · · · , e . We now multiply the whole
equation by the constant matrix P and we recall that P −1P = In, then
e −D(τ−t0)
−1
(P P )h(τ) dτ .
D(t−t0)
P y(t) = P e (P −1 P )y0 + (P −1 t
P )
t0
Recalling that P y = x, P y0 = x0, P h = g, and eAt = P eDtP −1 , we obtain
t
A(t−t0)
x(t) = e x0 +
t0
e −A(τ−t0) g(τ) dτ
This establishes the Theorem.
We said it in the homogenoeus equation and we now say it again. Although the expression
for the solution given in Eq. (5.6.8) looks simple, the exponentials of the coefficient matrix
are actually not that simple to compute. We show this in the following example.
Example 5.6.5: Find the vector-valued solution x to the differential system
x ′
3 1 2 1
= A x + g, x(0) = , A = , g = .
2 2 1 2
Solution: In Example 5.6.1 we have found the eigenvalues and eigenvectors of the coefficient
matrix, and the result is
λ1 = 3, v (1)
1
= ,
1
and λ2 = −1, v (2)
−1
= .
1
With this information and Theorem 5.5.7 we obtain that
and also that
so we conclude that
e At = 1
2
A = P DP −1 , P =
e At = P e Dt P −1 =
(e 3t + e −t ) (e 3t − e −t )
(e 3t − e −t ) (e 3t + e −t )
1 −1
1 1
, D =
1
3t −1 e 0
1 1 0 e−t
1
2
The solution to the initial value problem above is,
Since
e At x0 = 1
2
in a similar way
e −Aτ g = 1
2
⇒ e −At = 1
2
x(t) = e At x0 + e At
t
(e 3t + e −t ) (e 3t − e −t )
(e 3t − e −t ) (e 3t + e −t )
(e −3τ + e τ ) (e −3τ − e τ )
(e −3τ − e τ ) (e −3τ + e τ )
0
e −Aτ g dτ.
3
2
1
2
Integrating the last expresion above, we get
t
e −Aτ g dτ = 1
−3t t −e − e
2 −e−3t + et
+
Therefore, we get
x(t) = 1
2
5e 3t + e −t
5e 3t − e −t
0
+ 1
2
.
3 0
,
0 −1
1 1
,
−1 1
−3t t −3t t (e + e ) (e − e )
(e−3t − et ) (e−3t + et
.
)
= 1
2
= 1
2
5e 3t + e −t
5e 3t − e −t
3e −3τ − e τ
1
.
0
3t −t (e + e ) 3t −t (e − e )
(e3t − e−t ) (e3t + e−t 1
) 2
3e −3τ + e τ
,
.
−3t t −e − e
+
−e −3t + e t
1
.
0

G. NAGY – ODE July 2, 2013 205
Multiplying the matrix-vector product on the second term of the left-hand side above,
x(t) = 1
3t −t 5e + e
2 5e3t − e−t
1
− +
0
1
3t −t (e + e )
2 (e3t − e−t
.
)
We conclude that the solution to the initial value problem above is
x(t) =
3e 3t + e −t − 1
3e 3t − e −t
.
⊳

206 G. NAGY – ODE july 2, 2013
5.6.4. Exercises.
5.6.1.- . 5.6.2.- .

G. NAGY – ODE July 2, 2013 207
5.7. 2 × 2 Real distinct eigenvalues
This and the next two Sections are dedicated to study 2×2 linear, homogeneous differential
systems. In this Section we consider the case that the coefficient matrix is diagonalizable
and has two different, real, eigenvalues. We study in some depth the solutions of this particular
type of systems
5.7.1. Phase diagrams. The main tool will be the sketch of phase diagrams, also called
phase portraits. The solution value at a particular t is given by a 2-vector
x1(t)
x(t) = ,
x2(t)
so it can be represented by a point on a plane, while the solution function for all t ∈ R
corresponds to a curve on that plane. In the case that the solution vector x(t) is interpreted
as the position function of a particle moving in a plane at the time t, the curve given in
the phase diagram is the trajectory of the particle. Arrows are added to this trajectory to
indicate the motion of the particle as time increases.
In this Section we concentrate in constructing phase diagrams of solutions of 2×2 systems
x ′ (t) = Ax(t),
where matrix A has two real, distinct, eigenvalues. Theorem 5.5.8 implies that A is diagonalizable,
and we can use the results in Section 5.6 to describe the general solution of the
system, given by Eq. (5.6.5), that is,
x(t) = c1 v (1) e λ1t + c2 v (2) e λ2t . (5.7.1)
where λ1, v (1) and λ2, v (2) are eigenpairs of matrix A, with the eigenvectors forming a
linearly independent set.
In the case that one of the eigenvalues vanishes the study is simple and is left as an
exercise. In the case that the eigenvalues are non-zero, all possible phase diagrams can
be classified into three main classes, defined by the relative signs of the coefficient matrix
eigenvalues λ1 = λ2, as follows:
(i) 0 < λ 2 < λ 1, that is, both eigenvalues positive;
(ii) λ2 < 0 < λ1, that is, one eigenvalue negative and the other positive;
(iii) λ2 < λ1 < 0, that is, both eigenvalues negative.
A phase diagram contains several curves, each one corresponding to a solution given in
Eq. (5.7.1) for a particular choice of constants c 1 and c 2. A phase diagram can be sketched
following these simple steps:
(a) Plot the eigenvectors v (2) and v (1) corresponding to the eigenvalues λ 2 and λ 1.
(b) Draw the whole lines parallel to these vectors and passing through the origin. These
straight lines correspond to solutions with one of the coefficients c1 or c2 vanishing.
(c) Draw arrows on these lines indicate how the solution changes as the variable t grows.
If t is interpreted as time, the arrows indicate how the solution changes into the future.
The arrows point towards the origin if the corresponding eigenvalue λ is negative, and
they point away form the origin if the eigenvalue is positive.
(d) Find the non-straight curves correspond to solutions with both coefficient c 1 and c 2
non-zero. Again, arrows on these curves indicate the how the solution moves into the
future.
We now find the phase diagrams for three examples, one for every eigenvalue class above.

208 G. NAGY – ODE july 2, 2013
5.7.2. Case 0 < λ 2 < λ 1.
Example 5.7.1: Sketch the phase diagram of the solutions to the differential equation
x ′ = Ax, A = 1
11
4 1
3
.
9
(5.7.2)
Solution: The characteristic equation for this matrix A is given by
det(A − λI) = λ 2 − 5λ + 6 = 0 ⇒
λ1 = 3,
λ2 = 2.
One can show that the corresponding eigenvectors are given by
v (1)
3
= , v
1
(2)
−2
= .
2
So the general solution to the differential equation above is given by
x(t) = c1 v (1) e λ1t
+ c2 v (2) e λ2t
⇔
3
x(t) = c1
1
e 3t + c 2
−2
e
2
2t .
In Fig. 27 we have sketched four curves, each representing a solution x corresponding to a
particular choice of the constants c1 and c2. These curves actually represent eight different
solutions, for eight different choices of the constants c 1 and c 2, as is described below. The
arrows on these curves represent the change in the solution as the variable t grows. Since
both eigenvalues are positive, the length of the solution vector always increases as t grows.
The straight lines correspond to the following four solutions:
c1 = 1, c2 = 0, Line on the first quadrant, starting at the origin, parallel to v (1) ;
c 1 = 0, c 2 = 1, Line on the second quadrant, starting at the origin, parallel to v (2) ;
c1 = −1, c2 = 0, Line on the third quadrant, starting at the origin, parallel to −v (1) ;
c 1 = 0, c 2 = −1, Line on the fourth quadrant, starting at the origin, parallel to −v (2) .
2
v
x 2
Figure 27. The graph of several solutions to Eq. (5.7.2) corresponding to
the case 0 < λ2 < λ1, for different values of the constants c1 and c2. The
trivial solution x = 0 is called an unstable point.
v
1
x 1

G. NAGY – ODE July 2, 2013 209
Finally, the curved lines on each quadrant start at the origin, and they correspond to the
following choices of the constants:
c 1 > 0, c 2 > 0, Line starting on the second to the first quadrant;
c1 < 0, c2 > 0, Line starting on the second to the third quadrant;
c1 < 0, c2 < 0, Line starting on the fourth to the third quadrant,
c 1 > 0, c 2 < 0, Line starting on the fourth to the first quadrant.
5.7.3. Case λ 2 < 0 < λ 1.
Example 5.7.2: Sketch the phase diagram of the solutions to the differential equation
x ′ = Ax,
1
A =
3
3
.
1
(5.7.3)
Solution: In Example 5.6.3 we computed the eigenvalues and eigenvectors of the coefficient
matrix, and the result was
λ1 = 4, v (1)
1
= and λ2 = −2, v
1
(2)
−1
= .
1
In that Example we also computed the general solution to the differential equation above,
x(t) = c1 v (1) e λ1t
+ c2 v (2) e λ2t
⇔
1
x(t) = c1 e
1
4t
−1
+ c2 e
1
−2t ,
In Fig. 28 we have sketched four curves, each representing a solution x corresponding to a
particular choice of the constants c 1 and c 2. These curves actually represent eight different
solutions, for eight different choices of the constants c 1 and c 2, as is described below. The
arrows on these curves represent the change in the solution as the variable t grows. The
part of the solution with positive eigenvalue increases exponentially when t grows, while the
part of the solution with negative eigenvalue decreases exponentially when t grows. The
straight lines correspond to the following four solutions:
c1 = 1, c2 = 0, Line on the first quadrant, starting at the origin, parallel to v (1) ;
c 1 = 0, c 2 = 1, Line on the second quadrant, ending at the origin, parallel to v (2) ;
c1 = −1, c2 = 0, Line on the third quadrant, starting at the origin, parallel to −v (1) ;
c 1 = 0, c 2 = −1, Line on the fourth quadrant, ending at the origin, parallel to −v (2) .
Finally, the curved lines on each quadrant correspond to the following choices of the
constants:
c 1 > 0, c 2 > 0, Line from the second to the first quadrant,
c1 < 0, c2 > 0, Line from the second to the third quadrant,
c1 < 0, c2 < 0, Line from the fourth to the third quadrant,
c 1 > 0, c 2 < 0, Line from the fourth to the first quadrant.
⊳
⊳

210 G. NAGY – ODE july 2, 2013
v
1
2 v
−1
x
2
−1
Figure 28. The graph of several solutions to Eq. (5.7.3) corresponding to
the case λ2 < 0 < λ1, for different values of the constants c1 and c2. The
trivial solution x = 0 is called a saddle point.
5.7.4. Case λ2 < λ1 < 0.
Example 5.7.3: Sketch the phase diagram of the solutions to the differential equation
x ′ = Ax, A = 1
−9 3
. (5.7.4)
4 1 −11
Solution: The characteristic equation for this matrix A is given by
det(A − λI) = λ 2 + 5λ + 6 = 0 ⇒
λ1 = −2,
λ2 = −3.
One can show that the corresponding eigenvectors are given by
v (1)
3
= , v
1
(2)
−2
= .
2
So the general solution to the differential equation above is given by
x(t) = c1 v (1) e λ1t
+ c2 v (2) e λ2t
3
⇔ x(t) = c1 e
1
−2t + c2
1
1
x
1
−2
e
2
−3t .
In Fig. 29 we have sketched four curves, each representing a solution x corresponding to a
particular choice of the constants c1 and c2. These curves actually represent eight different
solutions, for eight different choices of the constants c 1 and c 2, as is described below.
The arrows on these curves represent the change in the solution as the variable t grows.
Since both eigenvalues are negative, the length of the solution vector always decreases as t
grows and the solution vector always approaches zero. The straight lines correspond to the
following four solutions:
c1 = 1, c2 = 0, Line on the first quadrant, ending at the origin, parallel to v (1) ;
c 1 = 0, c 2 = 1, Line on the second quadrant, ending at the origin, parallel to v (2) ;
c1 = −1, c2 = 0, Line on the third quadrant, ending at the origin, parallel to −v (1) ;
c 1 = 0, c 2 = −1, Line on the fourth quadrant, ending at the origin, parallel to −v (2) .

G. NAGY – ODE July 2, 2013 211
v
2
x 2
Figure 29. The graph of several solutions to Eq. (5.7.4) corresponding to
the case λ2 < λ1 < 0, for different values of the constants c1 and c2. The
trivial solution x = 0 is called a stable point.
Finally, the curved lines on each quadrant start at the origin, and they correspond to the
following choices of the constants:
c 1 > 0, c 2 > 0, Line entering the first from the second quadrant;
c1 < 0, c2 > 0, Line entering the third from the second quadrant;
c1 < 0, c2 < 0, Line entering the third from the fourth quadrant,
c 1 > 0, c 2 < 0, Line entering the first from the fourth quadrant.
v 1
x 1
⊳

212 G. NAGY – ODE july 2, 2013
5.7.5. Exercises.
5.7.1.- . 5.7.2.- .

G. NAGY – ODE July 2, 2013 213
5.8. 2 × 2 Complex eigenvalues
5.8.1. Complex conjugate pairs. In this Section we study the case that the coefficient
matrix is real-valued, diagonalizable, and has two complex-valued eigenvalues. Since the
coefficient matrix is real-valued, the complex-valued eigenvalues are always a complex conjugate
pair. The followng result holds for n × n matrices, not just 2 × 2.
Theorem 5.8.1 (Conjugate pairs). If an n × n real-valued matrix A has a complex
eigenvalue eigenvector pair λ, v, then the complex conjugate pair λ, v is an eigenvalue
eigenvector pair of matrix A.
Proof of Theorem 5.8.1: Complex conjugate the eigenvalue eigenvector equation for λ
and v, and recall that matrix A is real-valued, meaning that A = A, then we obtain,
Av = λv ⇔ Av = λ v,
This establishes the Theorem.
Since the complex eigenvalues of a matrix with real coefficients are always complex conjugate
pairs, there are an even number of complex eigenvalues and it holds that λ+ = λ−,
with a similar equation for their respective eigenvectors, that is, v (+) = v (−) . Therefore, it
is not difficult to see that an eigenvalue and eigenvector pair has the form
λ± = α ± iβ, v (±) = a ± ib, (5.8.1)
where α, β ∈ R and a, b ∈ R n .
It is simple to obtain a linearly independent set of two solutions to x ′ = Ax in the case
that matrix A has a complex conjugate pair of eigenvalues and eigenvectors. These solutions
are complex valued, but a simple linear combinantio of them is real-valued, as the following
result shows.
Theorem 5.8.2 (Complex and real solutions). If λ± = α ± iβ are two eigenvalues of
an n × n constant matrix A with respective eigenvectors v (±) = a ± ib, where α, β ∈ R and
a, b ∈ R n , and n 2, then the complex-valued functions
x (+) (t) = v (+) e λ+t , x (−) (t) = v (−) e λ-t , (5.8.2)
are solutions to the system x ′ = Ax and form a linearly independent set. Furthermore, the
real-valued functions
x (1) (t) = a cos(βt) − b sin(βt) e αt , x (2) (t) = a sin(βt) + b cos(βt) e αt . (5.8.3)
are also solutions to the system above and form a linearly independent set.
Proof of Theorem 5.8.2: We know from Theorem 5.6.2 that the two solutions in (5.8.2)
form a linearly independent set. The new information in Theorem 5.8.2 above is the realvalued
solutions in Eq. (5.8.3). They can be obtained from Eq. (5.8.2) as follows:
x (±) = (a ± ib) e (α±iβ)t
= e αt (a ± ib) e ±iβt
= e αt (a ± ib) cos(βt) ± i sin(βt)
= e αt a cos(βt) − b sin(βt) ± ie αt a sin(βt) + b cos(βt) .
Since the differential equation x ′ = Ax is linear, the functions below are also solutions,
x (1) = 1
(+) (−)
x + x
2
= e αt a cos(βt) − b sin(βt) ,
x (2) = 1 (+) (−)
x − x
2i
= e αt a sin(βt) + b cos(βt) .
This establishes the Theorem.

214 G. NAGY – ODE july 2, 2013
We now go back to 2 × 2 coefficient matrices.
Example 5.8.1: Find a real-valued set of fundamental solutions to the differential equation
x ′
2 3
= Ax, A = , (5.8.4)
−3 2
and then sketch a phase diagram for the solutions of this equation.
Solution: Fist find the eigenvalues of matrix A above,
0 = (2 − λ) 3
−3 (2 − λ) = (λ − 2)2 + 9 ⇒ λ± = 2 ± 3i.
Then find the respective eigenvectors. The one corresponding to λ+ is the solution of the
homogeneous linear system with coefficients given by
2 − (2 + 3i) 3 −3i 3 −i
=
→
−3 2 − (2 + 3i) −3 −3i −1
Therefore the eigenvector v
1 1
→
−i −1
i 1
→
−i 0
i
.
0
(+)
v1
= is given by
v1 = −iv2 ⇒ v2 = 1, v1 = −i, ⇒ v (+) =
v2
−i
, λ+ = 2 + 3i.
1
The second eigenvector is the complex conjugate of the eigenvector found above, that is,
v (−)
i
= , λ− = 2 − 3i.
1
Notice that
v (±) =
0
±
1
−1
i.
0
Then, the real and imaginary parts of the eigenvalues and of the eigenvectors are given by
0
−1
α = 2, β = 3, a = , b = .
1
0
So a real-valued expression for a fundamental set of solutions is given by
x (1)
=
0
−1
cos(3t) − sin(3t) e
1
0
2t ⇒ x (tione)
sin(3t)
= e
cos(3t)
2t ,
x (2)
=
0
−1
sin(3t) + cos(3t) e
1
0
2t ⇒ x (2)
− cos(3t)
=
e
sin(3t)
2t .
The phase diagram of these two fundamental solutions is given in Fig. 30 below. There
is also a circle given in that diagram, corresponding to the trajectory of the vectors
˜x (1)
sin(3t)
=
˜x
cos(3t)
(2)
− cos(3t)
=
.
sin(3t)
The trajectory of these vectors is a circle since their length is constant equal to one. ⊳
Remark: In the particular case that the coefficient matrix A in x ′ = Ax is 2 ×2, then there
are no other solutions linearly independent to the solutions given in Eq. (5.8.3). That is,
the general solution of given by
where
x(t) = c 1 x (1) (t) + c 2 x (2) (t),
x (1) = a cos(βt) − b sin(βt) e αt , x (2) = a sin(βt) + b cos(βt) e αt .

G. NAGY – ODE July 2, 2013 215
x (1)
b
x 2
a
Figure 30. The graph of the fundamental solutions x (1) and x (2) of the Eq. (5.8.4).
5.8.2. Qualitative analysis. We end this Section with a qualitative study of the phase
diagrams of the solutions for this case. We first fix the vectors a and b, and the plot phase
diagrams for solutions having α > 0, α = 0, and α < 0. These diagrams are given in Fig. 31.
We see that for α > 0 the solutions spiral outward as t increases, and for α < 0 the solutions
spiral inwards to the origin as t increases.
b
x 2
(2)
x
x (1)
a
x 1
b
(2)
x
x 2
Figure 31. The graph of the fundamental solutions x (1) and x (2) (dashed
line) of the Eq. (5.8.3) in the case of α > 0, α = 0, and α < 0, respectively.
Finally, let us study the following cases: We fix α > 0 and the vector a, and we plot the
phase diagrams for solutions with two choices of the vector b, as shown in Fig. 32. It can
then bee seen that the relative directions of the vectors a and b determines the rotation
direction of the solutions as t increases.
b
x 2
x
x (1)
(2)
a
x 1
Figure 32. The graph of the fundamental solutions x (1) and x (2) of the
Eq. (5.8.3) for α > 0, a given vector a, and for two choices of the vector b.
The relative positions of a and b determines the rotation direction.
x (1)
(1)
x
x
a
(2)
x 1
x1
x 2
x (1)
b
(2)
x
b
x 2
x
(2)
a
x 1
a
x 1

216 G. NAGY – ODE july 2, 2013
5.8.3. Exercises.
5.8.1.- . 5.8.2.- .

G. NAGY – ODE July 2, 2013 217
5.9. 2 × 2 Repeated eigenvalues
In Sections 5.7 and 5.8 we considered differential systems with coefficient matrix having
distinct eigenvalues. Theorem 5.5.8 implies that these matrices are diagonalizable. Hence,
the results in Section 5.6 can be applied to find the solutions to linear homogeneous systems
with such coefficient matrices. In this Section we study the remaining case, when the
coefficient matrix has repeated eigenvalues. We concentrate in the particular case where the
coefficient matrix has an eigenvalue with algebraic multiplicity r = 2, that is, the simplest
case. We have seen examples of matrices with r = 2 that are diagonalizable, Example 5.5.7,
and matrices that are not diagonalizable, Example 5.5.8. Both cases are possible, but only
the latter is new for us. Therefore we present a brief section where we summarize the main
result for diagonalizable systems with repeated eigenvalues, and then we concentrate in the
case of non-diagonalizable systems.
5.9.1. Diagonalizable systems. There is no new concepts in this case. If the n × n
coefficient matrix A of the linear system x ′ = Ax is diagonalizable, then the results in
Theorem 5.6.1 or Theorem 5.6.2 apply to this system. It does not matter whether matrix A
has repeated eigenvalues or not, as long as A has a linear independent set of n eigenvectors.
The following example illustrates this. The coefficient matrix has a repeated eigenvalue, but
since this matrix is diagonalizable, we use Theorem 5.6.2 to write down the general solution
of the differential system.
Example 5.9.1: Find a fundamental set of solutions to
x ′ ⎡
3 0
⎤
1
(t) = A x(t), A = ⎣0
3 2⎦
,
0 0 1
Solution: It is simple to see, and it was already computed in Example 5.5.7, that the
characteristic polynomial p(λ) = det(A − λI) is given by
p(λ) = −(λ − 3) 2 (λ − 1),
hence the eigenvalue λ1 = 3 has algebraic multiplicity r1 = 2, while the eigenvalue λ2 = 1
has algebraic multiplicity r2 = 1. We have also seen in Example 5.5.7 that the matrix A is
diagonalizable, with eigenvalues and eigenvectors
λ1 = 3,
⎡
⎣ 1
⎤ ⎡
0⎦
, ⎣
0
0
⎤
1⎦
0
and λ2 = 1,
⎡
⎣ −1
⎤
−2⎦
2
.
Since A is diagonalizable, we use Theorem 5.6.2 to write down a fundamental set of solution
to x ′ = Ax, and the result is,
⎡
x1(t) = ⎣ 1
⎤
0⎦
e
0
3t ⎡
, x2(t) = ⎣ 0
⎤
1⎦
e
0
3t ⎡
, x3(t) = ⎣ −1
⎤
−2⎦
e
2
t
.
5.9.2. Non-diagonalizable systems. Consider an n × n matrix A having an eigenvalue
λ with algebraic multiplicity r = 2, so λ is repeated, and geometric multiplicity s = 1, so
the eigenspace associated with λ has dimension one. A concrete example of this situation
was given in Example 5.5.8. If v is an eigenvector associated with λ, then a solution to the
differential system x ′ = Ax in this case is given by
x (1) (t) = v e λt .
⊳

218 G. NAGY – ODE july 2, 2013
However, every other eigenvector associated with λ is proportional to the v above. In other
words, every set of two eigenvectors associated with λ is linearly dependent. Then both
Theorem 5.6.1 and Theorem 5.6.2 do not hold for this system. The following result, stated
without a proof, provides a linearly independent set of two solutions to the system x ′ = Ax
associated with the repeated eigenvalue λ.
Theorem 5.9.1 (Repeated eigenvalue). If an n × n matrix A has an eigenvalue λ with
algebraic multiplicity r = 2 and only one associated eigen-direction, given by the eigenvector
v, then the differential system x ′ (t) = A x(t) has a linearly independent set of solutions
x (1) (t) = v e λt , x (2) (t) = v t + w e λt ,
where the vector w is one of infinitely many solutions of the algebraic linear system
(A − λI)w = v. (5.9.1)
Remark: We know that matrix (A − λI) is not invertible, since λ is an eigenvalue of A,
that is, det(A − λI) = 0. So an algebraic linear system with coefficient matrix (A − λI)
is not consistent for every source. Nevertheless, the Theorem above says that Eq. (5.9.1)
has solutions, and the property of v being an eigenvector of A is crucial to show that this
system is consistent. This proof is not reproduced here.
Example 5.9.2: Find the fundamental solutions of the differential equation
x ′ = Ax, A = 1
−6 4
.
4 −1 −2
Solution: As usual, we start finding the eigenvalues and eigenvectors of matrix A. The
former are the solutions of the characteristic equation
3
0 = − 2 − λ 1
− 1
1
4 − 2 − λ
=
λ + 3
λ +
2
1
+
2
1
4 = λ2 + 2λ + 1 = (λ + 1) 2 .
Therefore, there solution is
λ = −1, r = 2.
There is only one linearly independent eigenvector associated with this eigenvalue, and it is
computed as the solution of the linear system (A + I)v = 0, that is,
3 − 2 + 1 1
− 1
1
−
1 = 2 1
4 − 2 + 1 − 1
1 −2 1 −2
1 → →
⇒ v1 = 2v
4 2 1 −2 0 0
2.
Choosing v2 = 1, then v1 = 2, and we obtain
2
λ = −1, v = , r = 2, s = 1.
1
Following Theorem 5.9.1 we now must solve for the vector w the linear system (A+I)w = v.
The augmented matrix for this system is given by,
1 − 2 1
2 1 −2 −4
→ 1 −2
1 1 −2 →
−4 0 0
−4
0
⇒ w1 = 2w2 − 4.
− 1
4
1
2
We have obtained infinitely many solutions given by
2 −4
w = w2 + .
1 0
As one could have imagined, given any solution w, the cv + w is also a solution for any
c ∈ R. We choose the simplest solution given by
−4
w = .
0

G. NAGY – ODE July 2, 2013 219
Therefore, a fundamental set of solutions to the differential equation above is formed by
x (1)
2
(t) = e
1
−t , x (2)
(t) =
2 −4
t + e
1 0
−t . (5.9.2)
−x
(1)
x
(2)
w
x 2
Figure 33. The graph of the fundamental solutions x (1) , x (2) computed
in Example 5.9.2 and given in Eq. (5.9.2), together with their negatives.
The phase diagram of the fundamental solutions in Example 5.9.2 above are given in
Fig. 33. This diagram can be obtained as follows. First draw the vectors v and w. The
solution x (1) (t) is easy to draw, since it is represented by a straight line in the first quadrant
parallel to v ending in the origin. The solution approaches the origin as the variable t grows
since the eigenvalue λ = −1 is negative. The solution −x (1) is the the straight line in the
third quadrant ending in the origin and parallel to −v.
x (2)
−x
(1)
w
x 2
v
x
(1)
v
−x
(2)
x 1 x 1
−x (2)
(1)
−x
w
x
(2)
−x
(1)
x
1
x x 2
(2)
Figure 34. The trajectories of the functions x (1) , x (2) given in Eq. (5.9.3)
for the cases λ < 0 and λ > 0, respectively.
v
x (1)
⊳

220 G. NAGY – ODE july 2, 2013
The solution x (2) (t) is more difficult to draw. One way is to first draw the trajectory of
the time-dependent vector
˜x (2)
2 −4
= t + .
1 0
This is a straight line parallel to v passing through w, and is represented by a dashed line
in Fig. 33. As t grows the function ˜x (2) moves from the third quadrant to the second and
then to the first. The solution x (2) differs from ˜x (2) by the multiplicative factor e −t , so for
t < 0 we have that x (2) (t) has bigger length than ˜x (2) (t), while the opposite happens for
t > 0. In the limit t → ∞ the solution x (2) (t) approaches the origin, since the exponential
factor e −t decreases faster than the linear factor t grows. The result is then the upper half
of the s-shaped trajectory in Fig. 33. The lower half corresponds to −x (2) .
Given any vectors v and w, and any constant λ, the phase diagrams of functions
x (1) (t) = v e λt , x (2) (t) = v t + w e λt , (5.9.3)
can be obtained in a similar way as the one in Fig. 33. The results are displayed in Fig. 34
for the case λ < 0 and λ > 0, respectively.

5.9.3. Exercises.
5.9.1.- . 5.9.2.- .
G. NAGY – ODE July 2, 2013 221

222 G. NAGY – ODE july 2, 2013
Chapter 6. Boundary value problems
6.1. Eigenvalue-eigenfunction problems
In this Section we consider second order, linear, ordinary differential equations. In the
first half of the Section we study boundary value problems for these equations and in the
second half we focus on a particular type of boundary value problems, called the eigenvalueeigenfunction
problem for these equations.
6.1.1. Comparison: IVP and BVP. Given real constants a1 and a0, consider the second
order, linear, homogeneous, constant coefficients, ordinary differential equation
y ′′ + a 1 y ′ + a 0 y = 0. (6.1.1)
We now review the initial boundary value problem for the equation above, which was discussed
in Sect. 2.2, where we showed in Theorem 2.2.2 that this initial value problem always
has a unique solution.
Definition 6.1.1 (IVP). Given the constants t0, y0 and y1, find a solution y of Eq. (6.1.1)
satisfying the initial conditions
y(t 0) = y 0, y ′ (t 0) = y 1. (6.1.2)
There are other problems associated to the differential equation above. The following
one is called a boundary value problem.
Definition 6.1.2 (BVP). Given the constants t0 = t1, y0 and y1, find a solution y of
Eq. (6.1.1) satisfying the boundary conditions
y(t0) = y0, y(t1) = y1. (6.1.3)
One could say that the origins of the names “initial value problem” and “boundary value
problem” originates in physics. Newton’s second law of motion for a point particle was
the differential equation to solve in an initial value problem; the unknown function y was
interpreted as the position of the point particle; the independent variable t was interpreted
as time; and the additional conditions in Eq. (6.1.2) were interpreted as specifying the
position and velocity of the particle at an initial time. In a boundary value problem, the
differential equation was any equation describing a physical property of the system under
study, for example the temperature of a solid bar; the unknown function y represented any
physical property of the system, for example the temperature; the independent variable t
represented position in space, and it is usually denoted by x; and the additional conditions
given in Eq. (6.1.3) represent conditions on the physical quantity y at two different positions
in space given by t 0 and t 1, which are usually the boundaries of the system under study, for
example the temperature at the boundaries of the bar. This originates the name “boundary
value problem”.
We mentioned above that the initial value problem for Eq. (6.1.1) always has a unique
solution for every constants y0 and y1, result presented in Theorem 2.2.2. The case of the
boundary value problem for Eq. (6.1.1) is more complicated. A boundary value problem
may have a unique solution, or may have infinitely many solutions, or may have no solution,
depending on the boundary conditions. This result is stated in a precise way below.
Theorem 6.1.3 (BVP). Fix real constants a1, a0, and let r± be the roots of the characteristic
polynomial p(r) = r 2 + a 1r + a 0.
(i) If the roots r± ∈ R, then the boundary value problem given by Eqs. (6.1.1) and (6.1.3)
has a unique solution for all y 0, y 1 ∈ R.

G. NAGY – ODE July 2, 2013 223
(ii) If the roots r± form a complex conjugate pair, that is, r± = α±βi, with α, β ∈ R, then
the solution of the boundary value problem given by Eqs. (6.1.1) and (6.1.3) belongs
to only one of the following three possibilities:
(a) There exists a unique solution;
(b) There exists infinitely many solutions;
(c) There exists no solution.
Before presenting the proof of Theorem 6.1.3 concerning boundary value problem, let us
review part of the proof of Theorem 2.2.2 concerning initial value problems using matrix
notation to highlight the crucial part of the calculations. For the simplicity of this review,
we only consider the case r+ = r-. In this case the general solution of Eq. (6.1.1) can be
expressed as follows,
y(t) = c1 e r- t + c2 e r+ t , c1, c2 ∈ R.
The initial conditions in Eq. (6.1.2) determine the values of the constants c 1 and c 2 as
follows:
y0 = y(t0) = c1 e r- t0 r+ t0 + c2 e
y1 = y ′ (t0) = c1r- e r- t0 r+ t0 + c2r + e
⇒
r- t0 e r+ t0 e
r- er-t0 r+ t0 r + e
c1
=
c2 y0
.
y1 The linear system above has a unique solution c 1 and c 2 for every constants y 0 and y 1 iff
the determinant of the coefficient matrix Z is non-zero, where
A simple calculation shows
Z =
det(Z) = r+ − r-
r- t0 e r+ t0 e
r- er- t0 r+ t0 r+ e
.
e (r++r-) t0 = 0 ⇔ r+ = r-.
Since r+ = r-, the matrix Z is invertible and so the initial value problem above a unique
solution for every choice of the constants y 0 and y 1. The proof of Theorem 6.1.3 for the
boundary value problem follows the same steps we described above: First find the general
solution to the differential equation, second find whether the matrix Z above is invertible
or not.
Proof of Theorem 6.1.3:
Part (i): Assume that r± are real numbers. We have two cases, r+ = r- and r+ = r-. In
the former case the general solution to Eq. (6.1.1) is given by
y(t) = c1 e r- t + c2 e r+ t , c1, c2 ∈ R. (6.1.4)
The boundary conditions in Eq. (6.1.3) determine the values of the constants c1, c2, since
y0 = y(t0) = c1e r- t0
r+ t0
+ c2e r- t0 r+ t0
e e
⇒
er- t1
c1 y0
r+ t1 = . (6.1.5)
e c2 y1 y1 = y(t1) = c1e r- t1 r+ t1 + c2e
The linear system above has a unique solution c 1 and c 2 for every constants y 0 and y 1 iff
the determinant of the coefficient matrix Z is non-zero, where
r- t0 r+ t0
e e
Z =
. (6.1.6)
A straightforward calculation shows
er- t1 r+ t1 e
det(Z) = e r+ t1 e r- t0 − e r+ t0 e r- t1 = e r+ t0 e r- t0 e r+ (t1−t0) − e r- (t1−t0) . (6.1.7)
So it is simple to verify that
det(Z) = 0 ⇔ e r+ (t1−t0) = e r- (t1−t0) ⇔ r+ = r -. (6.1.8)

224 G. NAGY – ODE july 2, 2013
Therefore, in the case r + = r - the matrix Z is invertible and so the boundary value problem
above has a unique solution for every choice of the constants y0 and y1. In the case that
r+ = r- = r0, then we have to start over, since the general solution of Eq. (6.1.1) is not given
by Eq. (6.1.4) but by the following expression
y(t) = (c1 + c2 t) e r0t , c1, c2 ∈ R.
Again, the boundary conditions in Eq. (6.1.3) determine the values of the constants c1 and
c2 as follows:
y0 = y(t0) = c1e r0t0 + c2t0e r0t0
⇒
r0t0 e t0er0t0 er0t1 t1er0t1
c1 y0
= .
c2 y1 y 1 = y(t 1) = c 1e r0t1 + c2t 1e r0t1
The linear system above has a unique solution c1 and c2 for every constants y0 and y1 iff
the determinant of the coefficient matrix Z is non-zero, where
r0t0 e t0e
Z =
r0t0
A simple calculation shows
e r0t1 t1e r0t1
det(Z) = t1e r0(t1+t0) − t0e r0(t1+t0) = (t1 − t0)e r0(t1+t0) = 0 ⇔ t1 = t0.
Therefore, in the case r+ = r- = r0 the matrix Z is again invertible and so the boundary
value problem above has a unique solution for every choice of the constants y0 and y1. This
establishes part (i) of the Theorem.
Part (ii): Assume that the roots of the characteristic polynomial have the form r± = α±βi
with β = 0. In this case the general solution to Eq. (6.1.1) is still given by Eq. (6.1.4), and
the boundary condition of the problem are still given by Eq. (6.1.5). The determinant of
matrix Z introduced in Eq. (6.1.6) is still given by Eq. (6.1.7). However, the claim given
in Eq. (6.1.8) is true only in the case that r± are real numbers, and it does not hold in the
case that r± are complex numbers. The reason is the following calculation: Let us start
with Eq. (6.1.7) and then introduce that r± = α ± βi;
det(Z) = e (r++r-) t0 e r+ (t1−t0) − e r- (t1−t0)
= e 2αt0 e α(t1−t0) e iβ(t1−t0) − e −iβ(t1−t0)
= 2i e α(t1+t0) sin β(t1 − t 0) .
We then conclude that
det(Z) = 0 ⇔ sin β(t1 − t0) nπ
= 0 ⇔ β = , (6.1.9)
(t1 − t0)
where n = 1, 2, · · · and we are using that t1 = t0. This last equation in (6.1.9) is the key
to obtain all three cases in part (ii) of this Theorem, as can be seen from the following
argument:
Part (iia): If the coefficients a1, a0 in Eq. (6.1.1) and the numbers t1, t0 in the boundary
conditions in (6.1.3) are such that Eq. (6.1.9) does not hold, that is,
nπ
β =
(t1 − t0) ,
then det(Z) = 0 and so the boundary value problem for Eq. (6.1.1) has a unique solution
for all constants y 0 and y 1. This establishes part (iia).
Part (iib) and Part (iic): If the coefficients a1, a0 in Eq. (6.1.1) and the numbers t1, t0
in the boundary conditions in (6.1.3) are such that Eq. (6.1.9) holds, then det(Z) = 0, and
so the system of linear equation given in (6.1.5) may or may not have solutions, depending
on the values of the constants y0 and y1. In the case that there exists a solution, then there

G. NAGY – ODE July 2, 2013 225
are infinitely many solutions, since det(Z) = 0. This establishes part (iib). The remaining
case, when y0 and y1 are such that Eq. (6.1.5) has no solution is the case in part (iic). This
establishes the Theorem.
Our first example is a boundary value problem with a unique solution. This corresponds
to case (iia) in Theorem 6.1.3. The matrix Z defined in the proof of that Theorem is
invertible and the boundary value problem has a unique solution for every y 0 and y 1.
Example 6.1.1: Find the solution y(x) to the boundary value problem
y ′′ + 4y = 0, y(0) = 1, y(π/4) = −1.
Solution: We first find the general solution to the differential equation above. We know
that we have to look for solutions of the form y(x) = e rx , with the constant r being solutions
of the characteristic equation
r 2 + 4 = 0 ⇔ r± = ±2i.
We know that in this case we can express the general solution to the differential equation
above as follows,
y(x) = c1 cos(2x) + c2 sin(2x).
The boundary conditions imply the following system of linear equation for c1 and c2,
1 = y(0) = c1 1 0 c1 1
⇒
= .
−1 = y(π/4) = c2 0 1 c2 −1
The linear system above has the unique solution c1 = 1 and c2 = −1. Hence, the boundary
value problem above has the unique solution
y(x) = cos(2x) − sin(2x).
The following example is a small variation of the previous one, we change the value
of the constant t1, which is the place where we impose the second boundary condition,
from π/4 to π/2. This is enough to have a boundary value problem with infinitely many
solutions, corresponding to case (iib) in Theorem 6.1.3. The matrix Z in the proof of this
Theorem 6.1.3 is not invertible in this case, and the values of the constants t0, t1, y0, y1, a1
and a0 are such that there are infinitely many solutions.
Example 6.1.2: Find the solution y(x) to the boundary value problem
y ′′ + 4y = 0, y(0) = 1, y(π/2) = −1.
Solution: The general solution is the same as in Example 6.1.1 above, that is,
y(x) = c 1 cos(2x) + c 2 sin(2x).
The boundary conditions imply the following system of linear equation for c1 and c2,
1 = y(0) = c1 1 0 c1 1
⇒
= .
−1 = y(π/2) = −c1 −1 0 c2 −1
The linear system above has infinitely many solution, as can be seen from the following:
1 0 | 1 1 0 | 1 c1 = 1,
→
⇒
−1 0 | −1 0 0 | 0 c2 free.
Hence, the boundary value problem above has infinitely many solutions given by
y(x) = cos(2x) + c2 sin(2x), c2 ∈ R.
⊳
⊳

226 G. NAGY – ODE july 2, 2013
The following example again is a small variation of the previous one, this time we change
the value of the constant y1 from −1 to 1. This is enough to have a boundary value problem
with no solutions, corresponding to case (iic) in Theorem 6.1.3. The matrix Z in the proof
of this Theorem 6.1.3 is still not invertible, and the values of the constants t0, t1, y0, y1, a1
and a 0 are such that there is not solution.
Example 6.1.3: Find the solution y to the boundary value problem
y ′′ (x) + 4y(x) = 0, y(0) = 1, y(π/2) = 1.
Solution: The general solution is the same as in Examples 6.1.1 and 6.1.2 above, that is,
y(x) = c1 cos(2x) + c2 sin(2x).
The boundary conditions imply the following system of linear equation for c 1 and c 2,
1 = y(0) = c1
1 = y(π/2) = −c 1
From the equations above we see that there is no solution for c1, hence there is no solution
for the boundary value problem above.
Remark: We now use matrix notation, in order to follow the same steps we did in the
proof of Theorem 6.1.3:
1 0 c1
=
−1 0
c2
1
.
1
The linear system above has infinitely many solutions, as can be seen from the following
Gauss elimination operations
1 0
1 1 0 1
−1 0 → 1 0 1
1 0 0 →
2 0 0 1
Hence, there are no solutions to the linear system above. ⊳
6.1.2. Eigenvalue-eigenfunction problems. A particular type of boundary value problems
are called eigenvalue-eigenfunction problems. The main example we study in this
Section is the following: Find all the numbers λ and the non-zero functions with values y(x)
solutions of the homogeneous boundary value problem
y ′′ = λy, y(0) = 0, y(ℓ) = 0, ℓ > 0. (6.1.10)
This problem is analogous to the eigenvalue-eigenvector problem studied in Sect. 5.5, that is,
given an n × n matrix A find all numbers λ and non-zero vectors v solution of the algebraic
linear system Av = λv. The role of matrix A is played by d 2 /dx 2 , the role of the vector
space R n is played by the vector space of all infinitely differentiable functions f with domain
[0, ℓ] ⊂ R satisfying f(0) = f(ℓ) = 0. We mentioned in Sect. 5.5 that given any n × n matrix
A there exist at most n eigenvalues and eigenvectors. In the case of the boundary value
problem in Eq. (6.1.10) there exist infinitely many solutions λ and y(x), as can be seen in
the following result.
Theorem 6.1.4 (Eigenvalues-eigenfunctions). The homogeneous boundary value problem
in Eq. (6.1.10) has the infinitely many solutions, labeled by a subindex n ∈ N,
λn = − n2 π 2
ℓ2 , yn(x) = sin nπx
.
ℓ

G. NAGY – ODE July 2, 2013 227
Proof of Theorem 6.1.4: We first look for solutions having eigenvalue λ = 0. In such a
case the general solution to the differential equation in (6.1.10) is given by
y(x) = c1 + c2x, c1, c2 ∈ R.
The boundary conditions imply the following conditions on c1 and c2,
0 = y(0) = c1, 0 = y(ℓ) = c1 + c2ℓ
⇒ c1 = c2 = 0.
Since the only solution in this case is y = 0, there are no non-zero solutions.
We now look for solutions having eigenvalue λ > 0. In this case we redefine the eigenvalue
as λ = µ 2 , with µ > 0. The general solution to the differential equation in (6.1.10) is given
by
y(x) = c 1e −µx + c 2e µx ,
where we used that the roots of the characteristic polynomial r2 − µ 2 = 0 are given by
r± = ±µ. The boundary conditions imply the following conditions on c1 and c2,
0 = y(0) = c1 + c2,
1 1
⇒
e−µℓ e µℓ
c1 0
= .
0
Denoting by
we see that
0 = y(ℓ) = c1e −µℓ + c2e µℓ
Z =
1 1
e−µℓ e µℓ
det(Z) = e µℓ − e −µℓ = 0 ⇔ µ = 0.
Hence the matrix Z is invertible, and then we conclude that the linear system above for c 1,
c2 has a unique solution given by c1 = c2 = 0, and so y = 0. Therefore there are no non-zero
solutions y in the case that λ > 0.
We now study the last case, when the eigenvalue λ < 0. In this case we redefine the
eigenvalue as λ = −µ 2 , with µ > 0, and the general solution to the differential equation
in (6.1.10) is given by
y(x) = ˜c 1e −iµx + ˜c 2e iµx ,
where we used that the roots of the characteristic polynomial r 2 + µ 2 = 0 are given by
r± = ±iµ. In a case like this one, when the roots of the characteristic polynomial are
complex, it is convenient to express the general solution above as a linear combination of
real-valued functions,
y(x) = c 1 cos(µx) + c 2 sin(µx).
The boundary conditions imply the following conditions on c1 and c2,
0 = y(0) = c1, ⇒ c2 sin(µℓ) = 0.
0 = y(ℓ) = c1 cos(µℓ) + c2 sin(µℓ)
Since we are interested in non-zero solutions y, we look for solutions with c2 = 0. This
implies that µ cannot be arbitrary but must satisfy the equation
sin(µℓ) = 0 ⇔ µnℓ = nπ, n ∈ N.
We therefore conclude that the eigenvalues and eigenfunctions are given by
λn = − n2π2 ℓ2 , yn(x) = cn sin nπx
.
ℓ
Choosing the free constants cn = 1 we establish the Theorem.
c2

228 G. NAGY – ODE july 2, 2013
Example 6.1.4: Find the numbers λ and the non-zero functions with values y(x) solutions
of the following homogeneous boundary value problem
y ′′ = λy, y(0) = 0, y ′ (π) = 0.
Solution: This is also an eigenvalue-eigenfunction problem, the only difference with the
case studied in Theorem 6.1.4 is that the second boundary condition here involves the
derivative of the unknown function y. The solution is obtained following exactly the same
steps performed in the proof of Theorem 6.1.4.
We first look for solutions having eigenvalue λ = 0. In such a case the general solution
to the differential equation is given by
y(x) = c1 + c2x, c1, c2 ∈ R.
The boundary conditions imply the following conditions on c1 and c2,
0 = y(0) = c 1, 0 = y ′ (π) = c 2.
Since the only solution in this case is y = 0, there are no non-zero solutions with λ = 0.
We now look for solutions having eigenvalue λ > 0. In this case we redefine the eigenvalue
as λ = µ 2 , with µ > 0. The general solution to the differential equation is given by
y(x) = c1e −µx + c2e µx ,
where we used that the roots of the characteristic polynomial r2 − µ 2 = 0 are given by
r± = ±µ. The boundary conditions imply the following conditions on c1 and c2,
0 = y(0) = c1 + c2,
1 1
⇒
−µe−µπ µe µπ
c1 0
= .
c2 0
Denoting by
we see that
0 = y ′ (π) = −µc 1e −µπ + µc 2e µπ
Z =
1 1
−µe −µπ µe µπ
det(Z) = µ e µπ + e −µπ = 0.
Hence the matrix Z is invertible, and then we conclude that the linear system above for c 1,
c 2 has a unique solution given by c 1 = c 2 = 0, and so y = 0. Therefore there are no non-zero
solutions y in the case that λ > 0.
We now study the last case, when the eigenvalue λ < 0. In this case we redefine the
eigenvalue as λ = −µ 2 , with µ > 0, and the general solution to the differential equation
in (6.1.10) is given by
y(x) = ˜c1e −iµx + ˜c2e iµx ,
where we used that the roots of the characteristic polynomial r 2 + µ 2 = 0 are given by
r± = ±iµ. As we did in the proof of Theorem 6.1.4, it is convenient to express the general
solution above as a linear combination of real-valued functions,
y(x) = c1 cos(µx) + c2 sin(µx).
The boundary conditions imply the following conditions on c1 and c2,
0 = y(0) = c1, 0 = y ′
⇒ c2 cos(µπ) = 0.
(π) = −µc1 sin(µπ) + µc2 cos(µπ)
Since we are interested in non-zero solutions y, we look for solutions with c2 = 0. This
implies that µ cannot be arbitrary but must satisfy the equation
cos(µπ) = 0 ⇔ µnπ = (2n + 1) π
, n ∈ N.
2

G. NAGY – ODE July 2, 2013 229
We therefore conclude that the eigenvalues and eigenfunctions are given by
(2n + 1)2
λn = − ,
4
(2n + 1)x
yn(x) = cn sin
,
2
n ∈ N.
Example 6.1.5: Find the numbers λ and the non-zero functions with values y(x) solutions
of the homogeneous boundary value problem
x 2 y ′′ − x y ′ = λ y, y(1) = 0, y(ℓ) = 0, ℓ > 1.
Solution: This is also an eigenvalue-eigenfunction problem, the only difference with the
case studied in Theorem 6.1.4 is that the differential equation is now the Euler equation,
studied in Sect. 3.2, instead of a constant coefficient equation. Nevertheless, the solution is
obtained following exactly the same steps performed in the proof of Theorem 6.1.4.
Writing the differential equation above in the standard form of an Euler equation,
x 2 y ′′ − x y ′ − λy = 0,
we know that the general solution to the Euler equation is given by
y(x) = c 1 + c 2 ln(x) x r0
in the case that the constants r+ = r- = r0, where r± are the solutions of the Euler characteristic
equation
r(r − 1) − r − λ = 0 ⇒ r± = 1 ± √ 1 + λ.
In the case that r+ = r-, then the general solution to the Euler equation has the form
y(x) = c 1x r- + c 2x r+ .
Let us start with the first case, when the roots of the Euler characteristic polynomial are
repeated r+ = r- = r0. In our case this happens if 1 + λ = 0. In such a case r0 = 1, and the
general solution to the Euler equation is
y(x) = c1 + c2 ln(x) x.
The boundary conditions imply the following conditions on c1 and c2, 0 = y(1) = c1,
0 = y(ℓ) = c1 + c2 ln(ℓ)
⇒ c2ℓ ln(ℓ) = 0,
ℓ
hence c2 = 0. We conclude that the linear system above for c1, c2 has a unique solution
given by c1 = c2 = 0, and so y = 0. Therefore there are no non-zero solutions y in the case
that 1 + λ = 0.
We now look for solutions having eigenvalue λ satisfying the condition 1 + λ > 0. In this
case we redefine the eigenvalue as 1 + λ = µ 2 , with µ > 0. Then, r± = 1 ± µ, and so the
general solution to the differential equation is given by
y(x) = c 1x (1−µ) + c 2x (1+µ) ,
The boundary conditions imply the following conditions on c1 and c2,
0 = y(1) = c1 + c2,
0 = y(ℓ) = c1ℓ (1−µ) + c2ℓ (1+µ)
1 1
⇒
ℓ (1−µ) ℓ (1+µ)
c1
=
c2 Denoting by
Z =
1 1
ℓ (1−µ) ℓ (1+µ)
0
.
0
⊳

230 G. NAGY – ODE july 2, 2013
we see that
det(Z) = ℓ ℓ µ − ℓ −µ = 0 ⇔ ℓ = ±1.
Hence the matrix Z is invertible, and then we conclude that the linear system above for c1,
c 2 has a unique solution given by c 1 = c 2 = 0, and so y = 0. Therefore there are no non-zero
solutions y in the case that 1 + λ > 0.
We now study the second case, when the eigenvalue satisfies that 1 + λ < 0. In this case
we redefine the eigenvalue as 1 + λ = −µ 2 , with µ > 0. Then r± = 1 ± iµ, and the general
solution to the differential equation is given by
y(x) = ˜c 1x (1−iµ) + ˜c 2x (1+iµ) ,
As we did in the proof of Theorem 6.1.4, it is convenient to express the general solution
above as a linear combination of real-valued functions,
y(x) = x c1 cos µ ln(x) + c2 sin µ ln(x) .
The boundary conditions imply the following conditions on c1 and c2,
0 = y(1) = c1, 0 = y(ℓ) = c1ℓ cos µ ln(ℓ) + c2ℓ sin (µ ln(ℓ)
⇒ c2ℓ sin µ ln(ℓ) = 0.
Since we are interested in non-zero solutions y, we look for solutions with c 2 = 0. This
implies that µ cannot be arbitrary but must satisfy the equation
sin µ ln(ℓ) = 0 ⇔ µn ln(ℓ) = nπ, n ∈ N.
We therefore conclude that the eigenvalues and eigenfunctions are given by
λn = −1 − n2π2 ln 2 (ℓ) , yn(x)
nπ ln(x)
= cnx sin ,
ln(ℓ)
n ∈ N.
⊳

6.1.3. Exercises.
6.1.1.- . 6.1.2.- .
G. NAGY – ODE July 2, 2013 231

232 G. NAGY – ODE july 2, 2013
6.2. Overview of Fourier series
This Section is a brief introduction to the Fourier series expansions of periodic functions.
We first recall the origins of this type of series expansions. We then review basic few notions
of linear algebra that are satisfied by the set of infinitely differentiable functions. One crucial
concept is the orthogonality of the sine and cosine functions. We then introduce the Fourier
series of periodic functions, and we end this Section with the study two particular cases:
The Fourier series of odd and of even functions, which are called sine and cosine series,
respectively.
6.2.1. Origins of Fourier series. The study of solutions to the wave equation in one space
dimension by Daniel Bernoulli in the 1750s is a possible starting point to describe the origins
of the Fourier series. The physical system is a vibrating elastic string with fixed ends, the
unknown function with values u(t, x) represents the vertical displacement of a point in the
string at the time t and position x, as can be seen in the sketch given in Fig. 35. A constant
c > 0 characterizes the material that form the string.
y
0
x
u(t,x)
Figure 35. A sketch of a vibrating string in one space dimension that
moves only on the vertical direction and has fixed ends.
The mathematical problem to solve is the following initial-boundary value problem: Given
a function with values f(x) defined in the interval [0, ℓ] ⊂ R satisfying f(0) = f(ℓ) = 0, find
a function with values u(t, x) solution of the wave equation
∂ 2 t u(t, x) = c 2 ∂ 2 xu(t, x),
u(t, 0) = 0, u(t, ℓ) = 0,
u(0, x) = f(x), ∂tu(0, x) = 0.
The equations on the second line are called boundary conditions, since they are conditions
at the boundary of the vibrating string for all times. The equations on the third line are
called initial conditions, since they are equation that hold at the initial time only. The first
equation says that the initial position of the string is given by the function f, while the
second equation says that the initial velocity of the string vanishes. Bernoulli found that
the functions
cnπt
nπx
un(t, x) = cos sin
ℓ ℓ
are particular solutions to the problem above in the case that the initial position function
is given by
He also found that the function
u(t, x) =
nπx
fn(x) = sin .
ℓ
∞
cnπt
nπx
cn cos sin
ℓ ℓ
n=1
x

G. NAGY – ODE July 2, 2013 233
is also a solution to the problem above with initial condition
∞
nπx
f(x) = cn sin . (6.2.1)
ℓ
n=1
Is the set of initial functions f given in Eq. (6.2.1) big enough to include all continuous
functions satisfying the compatibility conditions f(0) = f(ℓ) = 0? Bernoulli said the answer
was yes, and his argument was that with infinitely many coefficients cn one can compute
every function f satisfying the compatibility conditions. Unfortunately this argument does
not prove Bernoulli’s claim. A proof would be a formula to compute the coefficients cn in
terms of the function f. However, Bernoulli could not find such a formula.
A formula was obtained by Joseph Fourier in the 1800s while studying a different problem.
He was looking for solutions to the following initial-boundary value problem: Given a
function with values f(x) defined in the interval [0, ℓ] ⊂ R satisfying f(0) = f(ℓ) = 0, and
given a positive constant k, find a function with values u(t, x) solution of the differential
equation
∂tu(t, x) = k ∂ 2 xu(t, x),
u(t, 0) = 0, u(t, ℓ) = 0,
u(0, x) = f(x).
The values of the unknown function u(t, x) are interpreted as the temperature of a solid
body at the time t and position x. The temperature in this problem does not depend on the
y and z coordinates. The partial differential equation on the first line above is called the
heat equation, and describes the variation of the body temperature. The thermal properties
of the body material are specified by the positive constant k, called the thermal diffusivity.
The main difference with the wave equation above is that only first time derivatives appear
in the equation. The boundary conditions on the second line say that both borders of the
body are held at constant temperature. The initial condition on the third line provides the
initial temperature of the body. Fourier found that the functions
nπ −(
un(t, x) = e ℓ )2kt sin
n=1
nπx
are particular solutions to the problem above in the case that the initial position function
is given by
nπx
fn(x) = sin .
ℓ
Fourier also found that the function
∞
nπ −(
u(t, x) = cn e ℓ )2
kt nπx
sin
ℓ
is also a solution to the problem above with initial condition
∞
nπx
f(x) = cn sin . (6.2.2)
ℓ
n=1
Fourier was able to show that any continuous function f defined on the domain [0, ℓ] ⊂ R
satisfying the conditions f(0) = f(ℓ) = 0 can be written as the series given in Eq. (6.2.2),
where the coefficients cn can be computed with the formula
cn = 2
ℓ
ℓ
0
f(x) sin
ℓ
nπx
dx.
ℓ

234 G. NAGY – ODE july 2, 2013
This formula for the coefficients cn, together with few other formulas that we will study
later on in this Section, was an important reason to name after Fourier instead of Bernoulli
the series containing those given in Eq. (6.2.2).
6.2.2. Fourier series. Every continuous τ-periodic function f can be expressed as an infinite
linear combination of sine and cosine functions. Before we present this result in a precise
form we need to introduce few definitions and to recall few concepts from linear algebra.
We start defining a peridic function saying that it is invariant under certain translations.
Definition 6.2.1. A function f : R → R is called τ-periodic iff for all x ∈ R holds
f(x − τ) = f(x), τ > 0.
The number τ is called the period of f, and the definition says that a function τ-periodic iff
it is invariant under translations by τ and so, under translations by any multiple of τ.
Example 6.2.1: The following functions are periodic, with period τ,
f(x) = sin(x), τ = 2π,
f(x) = cos(x), τ = 2π,
f(x) = tan(x), τ = π,
f(x) = sin(ax), τ = 2π
a .
The following function is also periodic and its graph is given in Fig. 36,
f(x) = e x , x ∈ [0, 2), f(x − 2) = f(x). (6.2.3)
y
y = f(x)
0 1 x
Figure 36. The graph of the function given in Eq. (6.2.3).
Example 6.2.2: Show that the following functions are τ-periodic for all n ∈ N,
2πnx
2πnx
fn(x) = cos , gn(x) = sin .
τ
τ
Solution: The following calculation shows that fn is τ-periodic,
2πn(x + τ)
fn(x + τ) = cos
,
τ
2πnx
= cos + 2πn ,
τ
2πnx
= cos ,
τ
= fn(x) ⇒ fn(xτ ) = fn(x).
A similar calculation shows that gn is τ-periodic. ⊳
⊳

G. NAGY – ODE July 2, 2013 235
It is simple to see that a linear combination of τ-periodic functions is also τ-periodic.
Indeed, let f and g be τ-periodic function, that is, f(x − τ) = f(x) and g(x − τ) = g(x).
Then, given any constants a, b holds
a f(x − τ) + b g(x − τ) = a f(x) + b g(x).
A simple application of this result is given in the following example.
Example 6.2.3: Show that the following function is τ-periodic,
f(x) = a0
2 +
∞
n=1
an cos
2πnx
τ
, +bn sin
2πnx
Solution: f is τ-periodic, since it is a linear combination of τ-periodic functions. ⊳
The set of infinitely many differentiable functions f : [−ℓ, ℓ] ⊂ R → R, with ℓ > 0,
together with the operation of linear combination of functions define a vector space. The
main difference between this vector space and Rn , for any n 1, is that the space of
functions is infinite dimensional. An inner product in a vector space is an operation that
associates to every pair of vectors a real number, and this operation is positive definite,
symmetric and bilinear. An inner product in the vector space of functions is defined as
follows: Given any two functions f and g, define its inner product, denoted by (f, g), as the
following number,
(f, g) =
ℓ
−ℓ
f(x) g(x) dx.
This is an inner product since it is positive definite,
it is symmetric,
(f, f) =
ℓ
−ℓ
(f, g) =
f 2 (x) dx 0, with
ℓ
−ℓ
f(x) g(x) dx =
and bilinear since it is symmetric and linear,
f, [ag + bh] =
= a
ℓ
ℓ
−ℓ
τ
.
(f, f) = 0 ⇔ f = 0 ;
g(x) f(x) dx = (g, f);
f(x) a g(x) + b h(x) dx
−ℓ
ℓ
f(x) g(x) dx + b ∈
−ℓ
ℓ −ℓ f(x) h(x) dx
= a(f, g) + b(f, h).
An inner product provides the notion of angle in a vector space, and so the notion of
orthogonality of vectors. The idea of perpendicular vectors in the three dimensional space
is indeed a notion that belongs to a vector space with an inner product, hence it can be
generalized to the space of functions. The following result states that certain sine and cosine
functions can be perpendicular to each oder.

236 G. NAGY – ODE july 2, 2013
Theorem 6.2.2 (Orthogonality). The following relations hold for all n, m ∈ N,
⎧
ℓ
nπx
mπx
⎪⎨
0 n = m,
cos cos dx = ℓ n = m = 0,
−ℓ ℓ ℓ ⎪⎩
2ℓ n = m = 0,
ℓ
nπx
mπx
0 n = m,
sin sin dx =
−ℓ ℓ ℓ
ℓ n = m,
ℓ
nπx
mπx
cos sin dx = 0.
ℓ ℓ
−ℓ
Remark: The proof of this result is based in the following trigonometric identities:
cos(θ) cos(φ) = 1
cos(θ + φ) + cos(θ − φ) ,
2
sin(θ) sin(φ) = 1
cos(θ − φ) − cos(θ + φ) ,
2
sin(θ) cos(φ) = 1
sin(θ + φ) + sin(θ − φ) .
2
Proof of Theorem 6.2.2: We show the proof of the first equation in Theorem 6.2.2, the
proof of the other two equations is similar. So, From the trigonometric identities above we
obtain
ℓ
nπx
mπx
cos cos dx =
−ℓ ℓ ℓ
1
ℓ
(n + m)πx
cos
dx +
2 −ℓ ℓ
1
ℓ
(n − m)πx
cos
dx.
2 −ℓ ℓ
Now, consider the case that at least one of n or m is strictly greater than zero. In this case
it holds the first term always vanishes, since
ℓ
1
(n + m)πx
ℓ
cos
dx =
2
ℓ
2(n + m)π sin
(n + m)πx
ℓ
= 0;
ℓ −ℓ
−ℓ
while the remaining term is zero in the sub-case n = m, due to the same argument as above,
ℓ
1
(n − m)πx
ℓ
cos
dx =
2
ℓ
2(n − m)π sin
(n − m)πx
ℓ
= 0;
ℓ −ℓ
−ℓ
while in the sub-case that n = m = 0 we have that
ℓ
1
(n − m)πx
cos
dx =
2
ℓ
1
2
−ℓ
ℓ
−ℓ
dx = ℓ.
Finally, in the case that both n = m = 0 is simple to see that
ℓ
nπx
mπx
ℓ
cos cos dx = dx = 2ℓ.
ℓ ℓ
−ℓ
This establishes the first equation in Theorem 6.2.2. The remaining equations are proven
in a similar way.
The main result of this Section is that any twice continuously differentiable function
defined on an interval [−ℓ, ℓ] ⊂ R, with ℓ > 0, admits a Fourier series expansion.
Theorem 6.2.3 (Fourier Series). Given ℓ > 0, assume that the functions f, f ′ f
and
′′ : [−ℓ, ℓ] ⊂ R → R are continuous. Then, f can be expressed as an infinite series
f(x) = a0
2 +
∞
nπx
nπx
an cos + bn sin
ℓ
ℓ
(6.2.4)
n=1
−ℓ

G. NAGY – ODE July 2, 2013 237
with the constants an and bn given by
an = 1
ℓ
nπx
f(x) cos dx, n 0, (6.2.5)
ℓ −ℓ
ℓ
bn = 1
ℓ
nπx
f(x) sin dx, n 1. (6.2.6)
ℓ
ℓ
−ℓ
Furthermore, the Fourier series in Eq. (6.2.4) provides a 2ℓ-periodic extension of f from
the domain [−ℓ, ℓ] ⊂ R to R.
Remark: The Fourier series given in Eq. (6.2.4) also exists for functions f which are only
piecewise continuous, although the proof of the Theorem for such functions is more involved
than in the case where f and its first two derivatives are continuous. In this notes we present
the simpler proof.
Proof of Theorem 6.2.3: We split the proof in two parts: First, we show that if f admits
an infinite series of the form given in Eq. (6.2.4), then the coefficients an and bn must
be given by Eqs. (6.2.5)-(6.2.6); second, we show that the functions fN approaches f as
N → ∞, where
fN (x) = a0
2 +
N
n=1
an cos
nπx
ℓ
+ bn sin
nπx
Regarding the first part, assume that f can be expressed by a series as given in Eq. (6.2.4).
Then, multiply both sides of this equation by a cosine function and integrate as follows,
ℓ
mπx
f(x) cos dx =
−ℓ
ℓ
a0
ℓ
mπx
cos dx
2 −ℓ ℓ
∞ ℓ
mπx
nπx
+ an cos cos dx
ℓ ℓ
n=1
ℓ
+ bn
−ℓ
−ℓ
ℓ
.
mπx
nπx
cos sin dx .
ℓ ℓ
In the case that m = 0 only the first term on the right-hand side above is nonzero, and
Theorem 6.2.2 implies that
ℓ
−ℓ
f(x) dx = a0
2 2ℓ ⇒ a0 = 1
ℓ
ℓ
−ℓ
f(x) dx,
which agrees with Eq. (6.2.5) for n = 0. In the case m 1 Theorem 6.2.2 implies that
ℓ
mπx
f(x) cos dx = anℓ,
ℓ
−ℓ
which again agrees with Eq. (6.2.5). Instead of multiplying by a cosine one multiplies the
equation above by sin
nπx
ℓ , then one obtains Eq. (6.2.6). The second part of the proof is
similar and it is left as an exercise. This establishes the Theorem.
Example 6.2.4: Find the Fourier series expansion of the function
1 + x x ∈ [−1, 0),
f(x) =
1 − x x ∈ [0, 1].
Solution: The Fourier series expansion is given by
f(x) = a0
2 +
∞
an cos(nπx) + bn sin(nπx) ,
n=1

238 G. NAGY – ODE july 2, 2013
where the coefficients an, bn are given in Theorem 6.2.3. We start computing a0, that is,
Similarly,
an =
=
Recalling the integrals
1
−1
0
a0 =
−1
=
1
−1
0
−1
f(x) dx
(1 + x) dx +
1
0
(1 − x) dx
= x + x2
0
+ x −
2 −1
x2
1
2 0
= 1 − 1
+
2
1 − 1
⇒ a0 = 1.
2
f(x) cos(nπx) dx
(1 + x) cos(nπx) dx +
1
cos(nπx) dx = 1
nπ sin(nπx),
x cos(nπx) dx = x
nπ
0
(1 − x) cos(nπx) dx.
1
sin(nπx) +
n2 cos(nπx),
π2 it is not difficult to see that
an = 1
nπ sin(nπx)
0
x
1
+ sin(nπx) +
−1 nπ n2 0
cos(nπx)
π2 −1
+ 1
nπ sin(nπx)
1
x
1
− sin(nπx) +
0 nπ n2 1
cos(nπx)
π2 0
1
=
n2 1
−
π2 n2
1
cos(−nπ) −
π2 n2 1
cos(−nπ) −
π2 n2π2
,
we then conclude that
an = 2
n2π2
1 − cos(−nπ) .
Finally, we must find the coefficients bn. The calculation is similar to the one done above
for an, and it is left as an exercise: Show that bn = 0. Then, the Fourier series of f is
f(x) = 1
2 +
∞ 2
n2π2
1 − cos(−nπ) cos(nπx).
n=1
6.2.3. Even and odd functions. There exist particular classes of functions with simple
Fourier series expansions. Simple means that either the bn or the an coefficients vanish.
These functions are called even or odd, respectively.
Definition 6.2.4. Given any ℓ > 0, a function f : [−ℓ, ℓ ] → R is called even iff holds
f(−x) = f(x), for all x ∈ [−ℓ, ℓ ].
A function f : [−ℓ, ℓ ] → R is called odd iff holds
f(−x) = −f(x), for all x ∈ [−ℓ, ℓ ].
⊳

Example 6.2.5: Even functions are the following:
G. NAGY – ODE July 2, 2013 239
f(x) = cos(x), f(x) = x 2 , f(x) = 3x 4 − 7x 2 + cos(3x).
Odd functions are the following:
f(x) = sin(x), f(x) = x 3 , f(x) = 3x 3 − 7x + sin(3x).
There exist functions that are neither even nor odd:
f(x) = e x , f(x) = x 2 + 2x − 1.
Notice that the product of two odd functions is even: For example f(x) = x sin(x) is even,
while both x and sin(x) are odd functions. Also notice that the product of an odd function
with an even function is again an odd function. ⊳
The Fourier series of a function which is either even or odd is simple to find.
Theorem 6.2.5 (Cosine and sine series). Consider the Fourier series of the function
f : [−ℓ, ℓ] → R, that is,
f(x) = a0
2 +
∞
nπx
nπx
an cos + bn sin .
ℓ
ℓ
n=1
(a) The function f is even iff the coefficients bn = 0 for all n 1. In this case the Fourier
series is called a cosine series.
(b) The function f is odd iff the coefficients an = 0 for all n 0. In this case the Fourier
series is called a sine series.
Proof of Theorem 6.2.5:
Part (a): Suppose that f is even, that is, f(−x) = f(x), and compute the coefficients bn,
ℓ
nπx
bn = f(x) sin dx
−ℓ
ℓ
−ℓ
nπ(−y)
= f(−y) sin (−dy), y = −x, dy = −dx,
ℓ
ℓ
ℓ
nπ(−y)
= f(−y) sin dy,
−ℓ
ℓ
ℓ
nπy
= − f(y) sin dy,
−ℓ
ℓ
= −bn ⇒ bn = 0.
Part (b): The proof is similar. This establishes the Theorem.

240 G. NAGY – ODE july 2, 2013
6.2.4. Exercises.
6.2.1.- . 6.2.2.- .

G. NAGY – ODE July 2, 2013 241
6.3. Application: The heat equation
In this Section we show how to find the solution to an initial-boundary value problem
for the heat equation. This equation is a partial differential equation, since the unknown
function u depends on two independent variables t and x, and the partial derivatives of
both variables appear into the equation. We find the solution to this problem solving
infinitely many initial value problems and boundary value problems for appropriate ordinary
differential equations.
6.3.1. The initial-boundary value problem. Consider the solid bar sketched in Fig. 37.
Assume that the bar is thermally insulated from the surrounding media except at the surfaces
x = 0 and x = ℓ, where the temperature is held constant to T0 and Tℓ, respectively.
This setting implies that the temperature in the bar does not depend on the coordinates
y and z, it depends only on the space coordinate x and time t. The thermal properties of
the bar material are characterized by its thermal diffusivity k > 0, which we assume to be
constant. Denoting by u(t, x) the bar temperature at the time t and the plane given by x
constant, it is known that the temperature must satisfy the heat equation
∂tu(t, x) = k ∂ 2 xu(t, x),
where ∂ denotes partial derivative. Notice that this is a partial differential equation for the
temperature function u.
T = 0
0
z
y
0 x L
insulation
insulation
T = 0
Figure 37. A solid bar thermally insulated on all surfaces except the x = 0
and x = L surfaces, which are held at temperatures T0 and TL, respectively.
The temperature of the bar is a function of the coordinate x only.
In this Section we are interested in solving an initial-boundary value problem for the
bar temperature: Fix positive constant ℓ and k, fix the constants T0 = Tℓ = 0, and fix an
infinitely differentiable function f : [0, ℓ] → R satisfying the conditions f(0) = f(ℓ) = 0.
Then, find a function u : [0, ∞) × [0, ℓ] → R, with values u(t, x), solution of the initialboundary
value problem,
∂tu(t, x) = k ∂ 2 xu(t, x), (6.3.1)
u(0, x) = f(x), (6.3.2)
u(t, 0) = 0, u(t, ℓ) = 0. (6.3.3)
The second line in the equation above is called the initial condition, while the equations on
the third line above are called the boundary conditions of the problem. A sketch on the tx
plane is useful to understand this type of problems, as can be seen in Fig. 38.
x
L

242 G. NAGY – ODE july 2, 2013
u ( t, 0 ) = 0
t
t
2
x
d u = k d u
0 u ( 0, x ) = f ( x ) L
u ( t, L ) = 0
Figure 38. A sketch on the tx plane of an initial-boundary value problem
for the heat equation.
6.3.2. Solution of the initial-boundary value problem. We use the method of separation
of variables to look for solutions to the Eqs. (6.3.1)-(6.3.3). The method of separation
of variables can be summarized in three main steps. First, look for a solution u as an infinite
series of the form u(t, x) = ∞ n=0 an un(t, x), where each function un is a solution of the
heat equation having a particularly simple form, in the sense that
un(t, x) = vn(t) wn(x).
The name of the method, separation of variables, originates in the equation above. The heat
equation for un implies particular ordinary differential equations for vn(t) and wn(x). So,
second, solve an appropriate initial value problem for each function vn and an appropriate
boundary value problem for each function wn, where the boundary conditions for wn are the
same as the boundary conditions for the function u. There exist infinitely many solutions
functions wn, which are mutually orthogonal. So, third, use the orthogonality relations
among the functions wn to find the coefficients an in terms of the initial data function f.
We then start looking for solutions of the form
∞
u(t, x) = an un(t, x), un(t, x) = vn(t) wn(x),
n=0
where each term is itself a solution of the heat equation, that is,
∂tun(t, x) − k ∂ 2 xun(t, x) = 0. (6.3.4)
Then it is clear that u will be solution of the heat equation, since
∂tu − k∂ 2 ∞
xu = an ∂tun − k∂ 2
xun = 0.
n=0
returning to Eq. (6.3.4) we see that this equation implies
wn(x) ∂tvn(t) = k vn(t)∂ 2 xwn(t) ⇒ 1 ∂tvn(t)
k vn(t) = ∂2 xwn(x)
wn(x) ,
where the last equation is obtained dividing the previous one by un. Since the left-hand
side on the last equation above depends only on t and the right-hand side depends only on
x, then any solution vn and wn must be such that each side is equal to a constant, that is,
1 ∂tvn(t)
k vn(t) = ∂2 xwn(x)
= λn,
wn(x)
where λn is a constant. The equation above contains two separate equations, given by
1 ∂tvn(t)
= λn,
k vn(t)
∂ 2 xwn(x)
wn(x)
x
= λn.

G. NAGY – ODE July 2, 2013 243
These equations are the motivation to propose the following problems for vn(t) and for
wn(x), respectively:
(i) First, find the function vn solution of initial value problem
v ′ n(t) = kλn vn(t) vn(0) = 1. (6.3.5)
(ii) Second, find all eigenvalues λn and all non-zero functions wn solutions of the boundary
value problem
w ′′
n(x) = λn wn(x), wn(0) = 0, wn(ℓ) = 0. (6.3.6)
The solution of Eq. (6.3.5) is straight forward to find, since it is a linear, first order, constant
coefficient equation. We have seen in Theorem 1.1.5 that the solution is given by
vn(t) = e kλnt .
The eigenvalue-eigenfunction problem in Eq. (6.3.6) was already studied in Sect. 6.1 and
the solution was found in Theorem 6.1.4. There is an infinite sequence of eigenvalues and
eigenfunctions given by
λn = − n2π2 ℓ2 , wn(x)
nπx
= sin , n ∈ N.
ℓ
With the functions vn and wn we can construct the function
u(t, x) =
∞
n=1
an e − n2π 2
ℓ2 kt
nπx
sin .
ℓ
We verify that this function satisfies the boundary conditions, since
∞
u(t, 0) = an e
n=1
− n2π 2
ℓ2 kt
nπ0
sin = 0,
ℓ
∞
u(t, ℓ) = an e − n2π 2
ℓ2 kt sin(nπ) = 0,
n=1
Therefore, the function u above is solution of Eqs. (6.3.1)-(6.3.3) iff it satisfies the initial
condition u(0, x) = f(x), that is,
∞
nπx
f(x) = an sin . (6.3.7)
ℓ
n=1
We have seen in Theorem 6.2.2 that the functions
nπx
∞ sin
ℓ
are mutually orthogonal on the interval [−ℓ, ℓ]. It is also not difficult to show that
⎧
ℓ
nπx
mπx
⎨ 0 n = m,
sin sin dx =
0 ℓ ℓ ⎩
ℓ
n = m,
2
Therefore the coefficients am can be obtained as follows: Multiply the Eq. (6.3.7) by the
function sin(mπx/ℓ) and integrate on the interval [0, ℓ]. Use the equation above to conclude
that
an = 2
ℓ
ℓ
0
n=1
nπx
f(x) sin dx.
ℓ

244 G. NAGY – ODE july 2, 2013
We therefore conclude that the solution to the initial-boundary value problem in Eqs. (6.3.1)-
(6.3.3) is given by
∞
u(t, x) = an e − n2π 2
ℓ2 kt
nπx
sin , an =
ℓ
2
ℓ
nπx
f(x) sin dx.
ℓ
ℓ
n=1
0

6.3.3. Exercises.
6.3.1.- . 6.3.2.- .
G. NAGY – ODE July 2, 2013 245

246 G. NAGY – ODE july 2, 2013
Chapter 1: First order equations
Chapter 7. Appendix
7.1. Review exercises
1.- . 2.- .

Chapter 2: Second order linear equations
1.- . 2.- .
G. NAGY – ODE July 2, 2013 247

248 G. NAGY – ODE july 2, 2013
Chapter 3: Power series solutions
1.- . 2.- .

Chapter 4: The Laplace Transform
1.- . 2.- .
G. NAGY – ODE July 2, 2013 249

250 G. NAGY – ODE july 2, 2013
Chapter 5: Systems of differential equations
1.- . 2.- .

Chapter 6: Boundary value problems
1.- . 2.- .
G. NAGY – ODE July 2, 2013 251

252 G. NAGY – ODE july 2, 2013
7.2. Practice Exams
Instructions to use the Practice Exams. The idea of these lecture notes is to help anyone
interested in learning linear algebra. More often than not such person is a college student.
An unfortunate aspect of our education system is that students must pass an exam to prove
they understood the ideas of a given subject. To prepare the student for such exam is the
purpose of the practice exams in this Section.
These practice exams once were actual exams taken by student in previous courses. These
exams can be useful to other students if they are used properly; one way is the following:
Study the course material first, do all the exercises at the end of every Section; do the review
problems in Section 7.1; then and only then take the first practice exam and do it. Think
of it as an actual exam. Do not look at the notes or any other literature. Do the whole
exam. Watch your time. You have only two hours to do it. After you finish, you can grade
yourself. You have the solutions to the exam at the end of the Chapter. Never, ever look at
the solutions before you finish the exam. If you do it, the practice exam is worthless. Really,
worthless; you will not have the solutions when you do the actual exam. The story does not
finish here. Pay close attention at the exercises you did wrong, if any. Go back to the class
material and do extra problems on those subjects. Review all subjects you think you are
not well prepared. Then take the second practice exam. Follow a similar procedure. Review
the subject related to the practice exam exercises you had difficulties to solve. Then, do the
next practice exam. You have three of them. After doing the last one, you should have a
good idea of what your actual grade in the class should be.

Practice Exam 1. (Two hours.)
1. .
G. NAGY – ODE July 2, 2013 253

254 G. NAGY – ODE july 2, 2013
Chapter 1: First order equations
7.3. Answers to exercises
Section 1.1: Linear constant coefficients equations
1.1.1.- It is simple to see that:
and that
y ′ = 2(t + 2) e 2t + e 2t ,
2y + e 2t = 2(t + 2) e 2t + e 2t .
Hence y ′ = 2y + e 2t . Also holds that
y(0) = (0 + 2) e 0 = 2.
Section 1.2: Linear variable coefficients equations
1.2.1.- y(t) = 4e −t − e −2t .
1.2.2.- y(t) = 2e t + 2(t − 1) e 2t .
1.2.3.- y(t) = π cos(t)
−
2t2 t2 .
1.2.4.- y(t) = c e (1+t2 ) 2
, with c ∈ R.
1.2.5.- y(t) = t2
n + 2
c
+ , with c ∈ R.
tn Section 1.3: Separable equations
1.3.1.- Implicit form: y2 t3
= + c.
2 3
2t3 Explicit form: y = ± + 2c.
3
1.3.2.- y 4 + y = t − t 3 + c, with c ∈ R.
1.3.3.- y(t) = 3
.
3 − t3 1.1.2.- y(t) = c e −4t + 1
, with c ∈ R.
2
1.1.3.- y(t) = 9
2 e−4t + 1
2 .
1.1.4.- y(x) = c e 6t − 1
, with c ∈ R.
6
1.1.5.- y(x) = 7
6 e6t − 1
6 .
1.1.6.- y(t) = 1
3 e3(t−1) + 2
3 .
1.2.6.- y(t) = c e t2
, with c ∈ R.
Let y1 and y2 be solutions, that is,
2ty1 − y ′ 1 = 0 and 2ty2 − y ′ 2 = 0. Then
2t(y1 + y2) − (y1 + y2) ′ =
(2ty1 − y ′ 1) + (2ty2 − y ′ 2) = 0.
1.2.7.- Define v(t) = 1/y(t). The equation
for v is v ′ = t v − t. Its solution is
v(t) = c e t2 /2
+ 1. Therefore,
1
y(t) =
t et2 .
/2 + 1
1.2.8.- y(x) = 6 + c e −x2 /4 2
√
1+t2 1.3.4.- y(t) = c e .
1.3.5.- y(t) = t ln(|t|) + c .
1.3.6.- y 2 (t) = 2t 2 ln(|t|) + c .
1.3.7.- Implicit: y 2 + ty − 2t = 0.
Explicit: y(t) = 1
−t + t2 + 8t .
2
1.3.8.- Hint: Recall the Definition
1.3.4 and use that
y ′ 1(x) = f x, y1(x) ,
for any independent variable x, for example
for x = kt.

Section 1.4: Exact equations
1.4.1.-
(a) The equation is exact. N = (1+t 2 ),
M = 2t y, so ∂tN = 2t = ∂yM.
(b) Since a potential function is given
by ψ(t, y) = t 2 y + y, the solution is
y(t) = c
t2 , c ∈ R.
+ 1
1.4.2.-
(a) The equation is exact. We have
N = t cos(y) − 2y, M = t + sin(y),
∂tN = cos(y) = ∂yM.
(b) Since a potential function is given
by ψ(t, y) = t2
2 + t sin(y) − y2 , the
solution is
t 2
2 + t sin(y(t)) − y2 (t) = c,
for c ∈ R.
G. NAGY – ODE July 2, 2013 255
1.4.3.-
(a) The equation is exact. We have
N = −2y + t e ty , M = 2 + y e ty ,
∂tN = (1 + t y) e ty = ∂yM.
(b) Since the potential function is given
by ψ(t, y) = 2t + e ty − y 2 , the solution
is
2t + e t y(t) − y 2 (t) = c,
for c ∈ R.
1.4.4.-
(a) µ(x) = 1/x.
(b) y 3 − 3xy + 18
5 x5 = 1.
1.4.5.-
(a) µ(x) = x 2 .
2
(b) y = − √ . The negative
1 + 2x4 square root is selected because the
the initial condition is y(0) < 0.

256 G. NAGY – ODE july 2, 2013
Section 1.5: Applications
1.5.1.-
(a) Denote m(t) the material mass as
function of time. Use m in mgr and
t in hours. Then
m(t) = m0 e −kt ,
where m0 = 50 mgr and k = ln(5)
hours.
(b) m(4) = 2
25 mgr.
(c) τ = ln(2)
hours, so τ 0.43 hours.
ln(5)
1.5.2.- Since
the condition
implies that
Q(t) = Q0e −(ro/V0)t ,
Q1 = Q0e −(ro/V0)t1
t1 = V0
ro
ln
Q0
Q1
.
Therefore, t = 20 ln(5) minutes.
1.5.3.- Since
−(ro/V0)t
Q(t) = V0qi 1 − e
and
lim Q(t) = V0qi,
t→∞
the result in this problem is
Q(t) = 300 1 − e −t/50
and
lim Q(t) = 300 grams.
t→∞
1.5.4.- Denoting ∆r = ri − ro and
V (t) = ∆r t + V0, we obtain
Q(t) =
+ qi
V0
V (t)
ro
∆r
Q0
V (t) − V0
V0
V (t)
ro
∆r
.
A reordering of terms gives
ro
V0 ∆r
Q(t) = qiV (t) − (qiV0 − Q0)
V (t)
and replacing the problem values yields
Q(t) = t + 200 − 100
(200) 2
.
(t + 200) 2
The concentration q(t) = Q(t)/V (t) is
q(t) = qi −
V0
V (t)
ro
∆r +1
qi − Q0
V0
The concentration at V (t) = Vm is
qm = qi −
V0
Vm
ro
∆r +1
qi − Q0
V0
.
,
which gives the value
qm = 121
125 grams/liter.
In the case of an unlimited capacity,
limt→∞ V (t) = ∞, thus the equation for
q(t) above says
lim q(t) = qi.
t→∞

Section 1.6: Non-linear equations
1.6.1.-
(a) Write the equation as
y ′ 2 ln(t)
= −
(t2 − 4) y.
The equation is not defined for
t = 0 t = ±2.
This provides the intervals
(−∞, −2), (−2, 2), (2, ∞).
Since the initial condition is at t =
1, the interval where the solution is
defined is
D = (0, 2).
(b) The equation is not defined for
t = 0, t = 3.
This provides the intervals
(−∞, 0), (0, 3), (3, ∞).
Since the initial condition is at t =
−1, the interval where the solution
is defined is
D = (−∞, 0).
G. NAGY – ODE July 2, 2013 257
1.6.2.-
(a) y = 2
3 t.
(b) Outside the disk t 2 + y 2 1.
1.6.3.-
(a) Since y = y2 0 − 4t2 , and the initial
condition is at t = 0, the solution
domain is
D = − y0
y0
, .
2 2
y0
(b) Since y =
1 − t2 and the initial
y0
condition is at t = 0, the solution
domain is
D = − 1
√ ,
y0
1
√ .
y0

258 G. NAGY – ODE july 2, 2013
Chapter 2: Second order linear equations
Section 2.1: Variable coefficients
2.1.1.- . 2.1.2.- .
Section 2.2: Constant coefficients
2.2.1.- . 2.2.2.- .
Section 2.3: Complex roots
2.3.1.- . 2.3.2.- .
Section 2.4: Repeated roots
2.4.1.- . 2.4.2.- .
Section 2.5: Undetermined coefficients
2.5.1.- . 2.5.2.- .
Section 2.6: Variation of parameters
2.6.1.- . 2.6.2.- .

Chapter 3: Power series solutions
Section 3.1: Regular points
3.1.1.- . 3.1.2.- .
Section 3.2: The Euler equation
3.2.1.- . 3.2.2.- .
Section 3.3: Regular-singular points
3.3.1.- . 3.3.2.- .
G. NAGY – ODE July 2, 2013 259

260 G. NAGY – ODE july 2, 2013
Chapter 4: The Laplace Transform
Section 4.1: Regular points
4.1.1.- . 4.1.2.- .
Section 4.2: The initial value problem
4.2.1.- . 4.2.2.- .
Section 4.3: Discontinuous sources
4.3.1.- . 4.3.2.- .
Section 4.4: Generalized sources
4.4.1.- . 4.4.2.- .
Section 4.5: Convolution solutions
4.5.1.- . 4.5.2.- .

Chapter 5: Systems of differential equations
Section 5.1: Introduction
5.1.1.- . 5.1.2.- .
Section 5.2: Systems of algebraic equations
5.2.1.- . 5.2.2.- .
Section 5.3: Matrix algebra
5.3.1.- . 5.3.2.- .
Section 5.4: Linear differential systems
5.4.1.- . 5.4.2.- .
Section 5.5: Diagonalizable matrices
5.5.1.- . 5.5.2.- .
Section 5.6: Constant coefficients systems
5.6.1.- . 5.6.2.- .
G. NAGY – ODE July 2, 2013 261

262 G. NAGY – ODE july 2, 2013
Chapter 6: Boundary value problems
Section 6.1: Eigenvalue-eigenfunction problems
6.1.1.- . 6.1.2.- .
Section 6.2: Overview of Fourier series
6.2.1.- . 6.2.2.- .
Section 6.3: Applications: The heat equation
6.3.1.- . 6.3.2.- .

G. NAGY – ODE July 2, 2013 263
References
[1] W. Boyce and R. DiPrima. Elementary differential equations and boundary value problems. Wiley, New
Jersey, 2009. 9th edition.
[2] W. Rudin. Principles of Mathematical Analysis. McGraw-Hill, New York, NY, 1953.