Math 225 Differential Equations Notes Chapter 2

Math 225 Differential Equations Notes Chapter 2
Michael Muscedere
September 26, 2004
1 Introduction
See Note Set 1
2 First-Order Differential Equations
2.1 Introduction: Motion of a Falling Body
We will revisit this subject later on.
2.2 Separable Equations
Definition If the right-hand side of the equation:
dy
= f(x, y)
dx
can be expressed as a function g(x) that depends only on x times a function
p(y) that depends only on y, then the differential equation is called
separable.
1

2.2.1 Method for Solving Separable Equations
Start with:
dy
dx = g(x)p(y)
multiply by dx and by h(y) := 1/p(y) to obtain
Then integrate both sides:
∫
h(y)dy = g(x)dx
∫
h(y)dy = g(x)dx
Obtain the implicit solution of the differential equation by merging together
the integration constants.
H(y) = G(x) + C
2

Example 1: Solve the non-linear equation
Solution:
1. y 2 dy = (x − 5)dx
2. ∫ y 2 dy = ∫ (x − 5)dx
dy
dx = x − 5
y 2
3. y3
3 = x2
2 − 5x + C
(
) 1/3
3x
4. y =
2
2 − 15x + 3C
(
) 1/3
3x
5. y =
2
2 − 15x + K 3

Example 2: Solve the Initial value problem
Solution:
• dy
y−1 = dx
x+3
• ∫ dy
y−1 = ∫ dx
x+3
• ln|y − 1| = ln|x + 3| + C
dy
dx = y − 1
x + 3
y(−1) = 0
• now we have an implicit solution
• Let’s solve for y explicitly
• e ln|y−1| = e ln|x+3|+C
• |y − 1| = e C |x + 3| = C 1 |x + 3| where C 1 = e C > 0
• y − 1 = ±C 1 (x + 3) implying y = 1 ± C 1 (x + 3)
• y = 1 + K(x + 3) If we let K := ±C 1 and non-zero
We can solve for K using the initial condition of y(−1) = 0
0 = 1 + K(−1 + 3) = 1 + 2K
So k = −1/2 and the solution to the IVP is
y = 1 − 1 2
−1
(x + 3) = (x + 1)
2
4

2.2.2 Formal Justification of Method
Let’s write the original DE in the following form.
h(y) dy
dx = g(x)
where h(y) := 1/p(y). Letting H(y) and G(x) denote antiderivatives
(indefinite integrals) of h(y) and g(x). so
H ′ (y) dy
dx = G′ (x)
By the chain rule the left hand side is the derivative of the composite
function H(y(x)) so
d
dx H(y(x)) = d
dx G(x)
Now, H(y(x)) and G(x) are two functions of x that have the same
derivative. Therefore they can only differ by a constant.
H(y(x)) = G(x) + C
Hence this is the implicit solution.
5

2.3 Linear Equations
Recall from Section 1.1 that a linear first-order equation is of the
form:
a 1 (x) dy
dx + a 0(x)y = b(x)
There are two situation for which the solution is quite immediate.
If a 0 (x) := 0 then
a 1 (x) dy
dx = b(x)
∫ b(x)
y(x) =
a 1 (x) dx + C
The second situation is if a 0 (x) happens to be equal to the derivative
of a 1 (x) In this case:
a 1 (x)y ′ + a 0 (x)y = a 1 (x)y ′ + a ′ 1(x)y = d
dx [a 1(x)y]
d
dx [a 1(x)y] = b(x)
∫
a 1 (x)y = b(x)dx + C
y(x) = 1 [∫
a 1 (x)
]
b(x)dx + C
6

A linear differential equation is seldom as easy to solve as the
first situation but we can always recast a linear DE into the second
situation by using integrating factors.
First let’s define a standard form.
dy
dx
+ P (x)y = Q(x)
where P (x) = a 0 (x)/a 1 (x) and Q(x) = b(x)a 1 (x)
Next we wish to find a function µ(x) such that if we multiply the
standard form by µ(x) then the left hand side becomes the derivative
of the product µ(x)y
7

Let’s see what µ(x) we need to make this happen.
µ(x) dy
dx
+ µ(x)P (x)y = µ(x)dy
dx + µ′ (x)y = d
dx µ(x)y
We see that if µ ′ (x) = µ(x)P (x) then we have situation two. So
• dµ(x)
dx
= µ(x)P (x) use separation of variables
• ∫ 1
µ dµ = ∫ P (x)dx
• µ(x) = e ∫ P (x)dx
We found the special µ(x) Yeh. so
d
µ(x)y = µ(x)Q(x)
dx
as before
µ(x)y =
∫
y(x) = 1 [∫
µ(x)
µ(x)Q(x)dx + C
]
µ(x)Q(x) + C
8

Example 3: Find the general solution to
1 dy
x dx − 2y
x = xcos(x)
2
• dy
dx − 2 x y = x2 cos(x) Put in standard form so P (x) = −2/x
• ∫ P (x)dx = ∫ −2
x
dx = −2ln|x| find µ(x)
• µ(x) = e −2ln|x| = e ln(x−2) = x −2
• x −2 dy
dx − 2x−3 y = cos(x) Multiply the standard form by µ(x)
• d
dx (x−2 y) = cos(x)
• x −2 y = ∫ cos(x)dx = sin(x) + C Integrate both sides
• so y = x 2 sin(x) + Cx 2 9

Example 4: Word Problem
A rock contains radioactive isotopes RA 1 and RA 2 that belong to
the same radioactive series:
• RA 1 decays into RA 2 at the rate of 50e −10t kg/sec.
• RA 2 decays into stable atoms at a rate proportional k to the
mass y(t) of RA 2 present
From this information we can setup a first order differential equation
describing the rate of change of RA 2 {y(t)}
dy
dt =
rate of creation - rate of decay
So
dy
dt = 50e−10t − ky(t)
If k = 2/sec and initially y(0) = 40 kg, find the mass y(t) of RA 2
for t ≥ 0
• dy
dt + 2y = 50e−10t Put in standard form so P (t) = 2
• ∫ P (t)dt = ∫ 2dx = 2t find µ(t)
• µ(t) = e 2t
• e 2t dy
dt + 2e2t y = 50e 2t e −10t Multiply the standard form by µ(t)
• d dt (e2t y) = 50e −8t
• e 2t y = ∫ 50e −8t dt Integrate both sides
• e 2t y = 50
−8 e−8t + C
• y = −25
4 e−10t + Ce −2t 10

To find C we need to apply the initial conditions y(0) = 40 to the
general solution.
40 = −25
4 e0 + Ce 0 = −25
4 + C
So C = 40 + 25/4 = 185/4
Thus the IVP solutions is:
y(t) = 185
4 e−2t − 25
4 e−10t
11

Theorem 2.1 Existence and Uniqueness Theorem
Suppose P (x) and Q(x) are continuous on an interval (a, b)
that contains the point x 0 . Then for any choice of initial values
y 0 there exist a unique solution y(x) on (a, b) to the initial value
problem
• dy
dx
+ P (x)y = Q(x)
• y(x 0 ) = y 0
In fact the solution is:
y(x) = 1
µ(x)
for the appropriate C.
[∫
]
µ(x)Q(x) + C
12

2.4 Exact Equations
It is also possible to use integrating factors to solve certain non-linear
differential equations. Consider an equation of the form
N(x, y) dy
dx
+ M(x, y) = 0
where M and N are arbitrary (differentiable) functions of x and y.
Multiplying through by the differential dx, we get
M(x, y)dx + N(x, y)dy = 0
Now suppose there exists a function F (x, y) with the following special
property.
∂F
∂x
= M(x, y)
∂F
∂y
= N(x, y)
then the preceding equation can be written in the form
∂F ∂F
dx +
∂x ∂y dy = 0
We recognize the left hand side as the total differential dF of a multivariable
function F so:
meaning that
dF = 0
F (x, y) = C
where C is a constant. Differential Equations where this is true are
called Exact
13

Example 5: Show the following equation is exact and give an
implicit solution.
(x + y) + (x + 3y) dy
dx = 0
We need to find an F such that:
∂F
∂x
= (x + y) and
∂F
∂y
= (x + 3y)
F (x, y) = (1/2)x 2 + xy + (3/2)y 2
Verify that this F does the trick.
So (1/2)x 2 + xy + (3/2)y 2 = C is the implicit solution.
14

Definition Exact Differential Form
The differential form M(x, y)dx + N(x, y)dy is said to be exact
differential form in a rectangle R if there is a function F (x, y) such
that:
∂F
∂x
= M(x, y)
∂F
∂y
total differential of F (x, y) satisfies:
= N(x, y) for all (x, y) in R. That is the
dF (x, y) ≡ ∂F ∂F
dx + dy = M(x, y)dx + N(x, y)dy
∂x ∂y
If M(x, y)dx + N(x, y)dy is an exact differential form, then the
equation
M(x, y)dx + N(x, y)dy = 0
is said to be an exact differential equation.
Before we waste time trying to find F (x, y) we need a test to see
if M(x, y)dx + N(x, y)dy is an exact differential form.
15

Theorem 2.2 Test for Exactness
Suppose the first partial derivatives of M(x, y) and N(x, y) are
continuous in a rectangle R. Then
M(x, y)dx + N(x, y)dy = 0
is an exact equation in R if and only if the compatibility condition
∂M
∂y
holds for all (x, y) in R.
Why is this true
∂N
(x, y) = (x, y)
∂x
Consider the continuous mixed partial derivatives of F (x, y). We
know from calculus that
∂ ∂F
∂y ∂x = ∂ ∂F
∂x ∂y
which leads directly to the compatibility condition.
16

2.4.1 Method for Solving Exact Equations
1. If M(x, y)dx + N(x, y)dy = 0 is exact, then ∂F/∂dx = M.
Integrate this last equation with respect to x to get
∫
F (x, y) = M(x, y)dx + g(y)
2. To determine g(y) take the partial derivative with respect to y
of both sides of the above integral equation and substitute N for
∂F/∂y. We can now solve for g ′ (y).
3. Integrate g ′ (y) to get g(y) up to a numerical constant. Substituting
g(y) into the integral equation above gives F (x, y)
4. The solution is F(x,y)=C
Alternatively, starting with ∂F/∂y = N, the implicit solution can
be found by first integrating with respect to y.
17

Example 6: Solve
(2xy − sec 2 x)dx + (x 2 + 2y)dy = 0
Solution:
Here M(x, y) = (2xy − sec 2 x) and N(x, y) = (x 2 + 2y) Testing for
exactness:
∂M ∂N
= 2x =
∂y ∂x
Integrate M(x, y) with respect to x
F (x, y) =
∫
(2xy − sec 2 x)dx + g(y) = x 2 y − tan(x) + g(y)
Next, take the partial derivative w.r.t y of the integral equation substituting
∂F
∂y = N(x, y) = (x2 + 2y)
x 2 + 2y = x 2 + g ′ (y)
so g ′ (y) = 2y implying g(y) = y 2 substituting in the equation for
F (x, y) we get
F (x, y) = x 2 y − tan(x) + g(y) = x 2 y − tan(x) + y 2
and the implicit solution of the DE is
x 2 y − tan(x) + y 2 = C
18

2.5 Exact Equations
Can we use integrating factors to turn DE into exact DE equations
Method for Finding Integrating Factors
1. If Mdx + Ndy = 0 is neither separable nor linear, compute
∂M ∂N ∂M
∂y
and
∂x
. If
∂y
= ∂N
∂x
then the DE is exact. use method in
previous section 2.4.
2. If not exact, consider
∂M/∂y − ∂N/∂x
N
If this is just a function of x then use the following as the integrating
factor.
[∫ ( ) ]
∂M/∂y − ∂N/∂x
µ(x) = exp
dx
N
3. If not, consider
∂N/∂x − ∂M/∂y
M
If this is just a function of y then use the following as the integrating
factor.
[∫ ( ) ]
∂N/∂x − ∂M/∂y
µ(y) = exp
dy
M
Example 7: Solve
Solution:
Test for exactness:
(2x 2 + y)dx + (x 2 y − x)dy = 0
∂M
∂y
= 1 ≠ (2xy − 1) =
∂N
∂x
19

So compute
∂M/∂y − ∂N/∂x
N
=
1 − (2xy − 1)
x 2 y − x
=
2(1 − xy)
−x(1 − xy = −2
x
This is only a function of x so the integrating factor is:
(∫ ) −2
µ(x) = exp
x dx = x −2
multiplying the differential equation by µ we get
(2 + yx −2 )dx + (y − x −1 )dy = 0
∂M
∂y = x−2 = ∂N
∂x
This equation is now exact. Using the method of 2.4 we get the
implicit solution:
2x − yx −1 + y2
2 = C
2.6 Substitution and Transformations
Substitution Procedure
1. Identify the appropriate substitution for the given problem type.
2. Rewrite the original equation in terms of the new variables
3. Solve the transformed equation
4. Express the solution in terms of the original variables. Noting
any solutions lost in the procedure.
20

Definition Homogeneous Equations
If the right hand side of the equation
dy
= f(x, y)
dx
can be expressed as a function of the ratio y/x then we say the
equation is homogeneous.
One test to see if the DE in the definition is homogeneous is to replace
x with tx and y with ty then the equation is homogeneous if and
only if
f(tx, ty) = f(x, y)
for all t ≠ 0
To solve a homogeneous equation, we can make the following substitutions
and by the chain rule on y = vx
Example 8: Solve
v = y x
dy
dx = v d
dx x + xdv dx = v + xdv dx
Solution:
(xy + y 2 + x 2 )dx − x 2 dy = 0
• Is the equation exact
No
∂M
∂y
= x + 2y ≠ 2x =
∂N
∂x
21

• Is the equation linear
No
• Is it a function of y/x
dy
dx = xy + y2 + x 2
x 2
dy
dx = xy + y2 + x 2
x 2 = y x + (y x )2 + 1
So the equation is homogeneous.
using the substitution we get:
v + x dv
dx = v + v2 + 1
This equation is separable so:
• ∫ ∫ 1 1
v1
2 dv =
x dx
•
•
•
•
arctan(v) = ln|x| + C
v = tan(ln|x| + C)
y
x
= tan(ln|x| + C)
y = xtan(ln|x| + C)
• Also note x ≡ 0 is also a solution.
22

2.6.1 Equations of the form dy
dx
If the right side of the equation:
= G(ax + by)
dy
= f(x, y)
dx
can be expressed as a function of ax + by, That is
then the substitution
dy
dx
= G(ax + by)
z = ax + by
transforms the equation into a separable one.
Example 9: Solve
Solution:
dy
dx
= y − x − 1 + (x − y + 2)−1
•
dy
dx = y − x − 1 + (x − y + 2)−1 = −(x − y) − 1 + [(x − y) + 2] −1
•
z = x − y
•
dz
dx = 1 − dy
dx
•
1 − dz = −z − 1 + [z + 2]−1
dx
•
dz
= (z + 2) − (z + 2)−1
dx
This equation is separable.
23

• ∫
• ∫
•
•
•
replacing z by x − y
∫
dz
(z + 2) − (z + 2) = −1
∫
(z + 2)dz
(z + 2) 2 − 1 =
dx
1
2 ln|(z + 2)2 − 1| = x + C 1
(z + 2) 2 = Ce 2x + 1
(x − y + 2) 2 = Ce 2x + 1
dx
2.6.2 Bernoulli Equations
Definition Bernoulli Equations
A first order equation that can be written in the form
dy
+ P (x)y = Q(x)yn
dx
where P(x) and Q(x) are continuous on an interval (a,b) and n is a
real number is called a Bernoulli Equation.
Note that when n=0 or n=1 the equation is linear and can be solved
by our known methods.
24

To solve Bernoulli equations, we can make the following substitutions
v = y 1−n
suppose we divide Bernoulli’s equation by y n then:
y −ndy
dx +P (x)y1−n = Q(x) Eq 1
Notice that from the chain rule applied to v = y 1−n
dv
= (1 − n)y−ndy
dx dx
So substituting into Eq 1. we get
1 dv
+ P (x)v = Q(x)
1 − n dx
This last equation is linear and can be solved using integrating factors.
25

Example 10: Solve
dy
dx − 5y = −5 2 xy3
Solution:
This is a Bernoulli equation with n = 3, P (x) = −5, and Q(x) =
−5x/2 So as described before divide by y 3
y −3dy
dx − 5y−2 = − 5 2 x
Now make the substitution v = y 1−3 = y −2 and dv/dx = −2y −3 dy/dx
− 1 dv
2 dx − 5v = −5 2 x
dv
+ 10v = 5x
dx
So now we can use integrating factors to get.
Substituting v = y −2 we get
v = x 2 − 1 20 + Ce−10x
y −2 = x 2 − 1 20 + Ce−10x
How about the solution y = 0 for all x
26