Second Order Differential Equations
Equivalent Forms; Initial Value Problems In general,
second order differential equations take the form
dx2 x, y, dy
where f(x, y, z) is a continuous function of the variables x, y
and z in some region R ⊂ R 3 and is continuously differentiable
with respect to y and z throughout that region.
If we define z = dy
, then we can write the above equation
equivalently as a system of two (coupled) first order equations:
= f(x, y, z).
Written in this form and drawing on experience with first order
equations, we would expect to have to give initial values
y(x0) = y0, z(x0) = z0,
in order to specify an initial value problem. Going back to the
second order equation, this means we need to give
y(x0) = y0, dy
dx (x0) = z0; equivalently : dy
dx (x0) = y1,
in order to specify an initial value problem for it.
General Solutions We say that an expression y(x, c1, c2) is a
general solution for the second order differential equation above
if, as a function of x, it satisfies the differential equation, and,
given initial values as above at a point x0, with (x0, y0, y1) in
the region R where f(x, y, z) has the required properties, we can
solve the equations
y(x0, c1, c2) = y0, dy
dx (x0, c1, c2) = y1
for c1 and c2 to satisfy that initial value problem.
Example 1 Consider the second order differential equation
in the region x > 0, and let us try to find solutions of the form
y = x r . Substituting, we have
r(r − 1) x r−2 = 2r x r−2 − 2x r−2
x r−2 (r 2 − 3r + 2) = 0.
Since x r−2 is not identically zero, we need to solve the so-called
r 2 − 3r + 2 = 0 −→ r = 2 or r = 1.
So we conclude that y(x) = x and y(x) = x 2 are solutions,
which is easily checked out. Since the equation is linear with
respect to y (if y and ˆy are solutions, so is any linear combination
of the two) it then makes sense to try a general solution of the
y(x, c1, c2) = c1 x + c2 x 2 .
To satisfy initial conditions as specified above at x0 > 0 we
c1 x0 + c2 x0 2 = y0
c1 + 2 c2 x0 = y1
a system of two linear algebraic equations. The determinant of
the coefficients of c1 and c2 on the left hand side is 2 x0 2 − x0 2 =
x0 2 = 0, so we conclude we can always obtain a unique solution
of these equations. We conclude therefore that y(x, c1, c2) =
c1 x + c2 x 2 is the general solution of the second order differential
equation d2y dx2 = 2
x 2y for x > 0.
A General Method of Solution in the “Autonomous”
If our second order differential equation does not depend explicitly
on x, which is technically referred to as the autonomous,
i.e., ”self-governing”, case, we can employ a transformation of
variables to reduce the second order equation to two first order
differential equations. Assuming the equation is
dx2 y, dy
Substituting this into the second derivative we have
= z dz
and thus the original second order equation becomes
= f(y, z).
If we can solve this last equation to get z = z(y, c1), then the
equation used to define z becomes
= z(y, c1)
which, being in separable form
can be integrated to give
dy = dx
dη = x + c2.
Then, at least in principle, we can solve this to get y = y(x, c1, c2).
Success of the project ultimately depends on being able to get a
closed form solution for the transformed second order equation,
z dz = f(y, z), which has now become a first order equation in
z and y.
Example 2 We consider the autonomous second order differential
dx2 dx .
Following the plan indicated above, we arrive at
= y z or dz
This is solved immediately to obtain
z = z(y, c1) = y2 + c1
Then we have
dx = y2 + c1
There are three cases here, depending on whether c1 is 0, positive,
or negative. Taking just the case where c1 is positive, we
write c1 = c2 and we have
c2 + y2dy = dx.
Setting y = c w we have
and we integrate to obtain
1 + w2
dw = dx
c tan−1 (w) = x + c2.
Then w = tan( c
2 (x + c2)) and we have as the general solution
y = y(x, c, c2) = c tan( c
(x + c2)).
The cases corresponding to c1 = 0 and c1 < 0 are handled in
a similar way but different expressions are obtained. (Try those
cases as an exercise.)
Linear Second Order Differential Equations
It would be misleading to say that we can solve most second
order differential equations. In fact, the ones we can solve are
rather special. Prominent among these are the linear second
order differential equations
+ q(x) y = g(x),
where p(x), q(x) and g(x) are known, continuous (or piecewise
continuous) functions of x. Such a differential equation is called
homogeneous if g(x) ≡ 0 and inhomogeneous if g(x) ≡ 0, just
as in the first order case. For the present we consider only the
+ q(x) y = 0.
A fundamental property of solutions of such a differential equation
is expressed in the following
Proposition If y1(x) and y2(x) are any two solutions of the
above linear homogeneous second order o.d.e., then for any constants
c1 and c2, the linear combination y(x) = c1 y1(x) +
c2 y2(x) is also a solution.
The proof is immediate when we use
c1 y1(x) + c2 y2(x)
(x) + c2
and the similar relationship for the second derivative of the linear
Constant Coefficient Linear Homogeneous Second Order
Differential Equations These are differential equations
of the form
+ p + q y = 0
where p and q are constants. These differential equations are
particularly easy to solve. Since in the linear homogeneous first
order constant coefficient case
+ p y = 0
we have the exponential solution y(x) = e −px it makes sense
to guess that the corresponding second order equation might
also have solutions of the form y(x) = e rx . Substituting this
proposed solution form into the differential equation we have
r 2 e rx + p r e rx + q e rx = e rx (r 2 + p r + q) = 0.
The exponential is never equal to zero so we must have
r 2 + p r + q = 0,
an algebraic equation which is called the characteristic equation
for the differential equation in question. Use of the quadratic
formula gives the two roots
r = −p ± √ p2 − 4 q
we will refer to the two roots thus obtained as r1 and r2. Then,
retracing our steps, we can see that with y(x) = e ri x , i = 1, 2,
dx2 dx + q y = 0 = eri x
(ri 2 + p ri + q) = 0, i = 1, 2,
so that these are, indeed, solutions. Then, from the Proposition
we have solutions
y(x, c1, c2) = c1 e r1 x + c2 e r2 x .
It is natural to suspect that this might be the general solution.
To check on this we suppose that we want to solve
y(x0, c1, c2) = y0
c1, c2) = y1
Substituting y(x, c1, c2) = c1 e r1 x + c2 e r2 x we have
c1 e r1 x0 + c2 e r2 x0 = y0
c1 r1 e r1 x0 + c2 r2 e r2 x0 = y1
The coefficients of the unknowns c1 and c2 form the array
This array has determinant
e r1 x0 e r2 x0
r1 e r1 x0 r2 e r2 x0
e (r1+r2)x0 (r2 − r1)
which is different from 0 if r1 = r2, i.e., if the roots of the
characteristic equation are distinct. Thus we have the form of
the general solution in this case. We postpone consideration of
the case r1 = r2, wherein r 2 + p r + q = (r − r1) 2 until a later
Example 3 Find the solution of the initial value problem
dt2 dt + 12 y = 0, y(0) = 0, y′ (0) = 1.
Solution In this case the characteristic equation is
r 2 − 7 r + 12 = (r − 3)(r − 4) = 0,
yielding the roots r1 = 3, r2 = 4. Thus the general solution is
Its derivative is
At t = 0 we require
y(t, c1, c2) = c1 e 3t + c2 e 4t .
y ′ (t, c1, c2) = 3 c1 e 3t + 4 c2 e 4t .
y(0, c1, c2) = c1 e 3·0 + c2 e 4·0 = c1 + c2 = 0,
y ′ (0, c1, c2) = 3 c1 e 3·0 + 4 c2 e 4·0 = 3 c1 + 4 c2 = 1.
Multiplying the first equation by 3 and subtracting the result
from the second equation we obtain c2 = 1. Then the first
equation gives c1 = − 1 and the desired solution is
y(t) = − e 3t + e 4t .
Example 4 Find the general solution of
6 d2y dy
Here the characteristic equation is
+ 15 y = 0.
6 r 2 − 19 r + 15 = 0.
The roots are not immediately obvious. We recall the quadratic
formula for finding the roots of a quadratic equation a r2 + b r +
c = 0:
r = −b ± √ b2 − 4 a c
Applying this formula to our case we obtain the roots
(19) 2 − 4 · 6 · 15
2 · 6
which yields the roots
r1 = 20
Thus the general solution is
3 ; r2 = 18
= 19 ± √ 361 − 360
y(t, c1, c2) = c1 e 5t/3 + c2 e 3t/2 .
Example 5 Two blocks of material are placed in contact
with each other on a large stone table whose temperature is
maintained at 0 ◦ . The heat transfer coefficient between the two
blocks and between the blocks and the table is k > 0. Describe
the evolution of the temperature T1(t) in the first block.
Solution Applying the laws of heat conduction we obtained
a coupled system of two first order equations:
dt = − k (T1 − T2) − k (T1 − 0) = − 2 k T1 + k T2; (I)
dt = − k (T2 − T1) − k (T2 − 0) = − 2 k T2 + k T1. (II)
We differentiate the equation (I) with respect to t:
= − 2 k
+ k . (III)
We substitute equations (I) and (II) into (III) and obtain
d 2 T1
dt 2 = − 2k ( − 2k T1 + k T2) + k ( − 2k T2 + k T1)
= 5 k 2 T1 − 4 k 2 T2. (IV)
To eliminate T2 in equation (IV) we multiply equation (I) by 4 k
dt = − 8 k2 T1 + 4 k 2 T2. (V)
Then adding equations (IV) and (V) we obtain
dt2 dt = − 3 k2 T1,
d 2 T1
dt + 3 k2 T1 = 0.
Here the characteristic equation is
with the roots
r 2 + 4k r + 3 k 2 = (r + k)(r + 3k) = 0
r1 = − k, r2 = − 3k.
Accordingly, th general solution is
T1(t, c1, c2) = c1 e −kt + c2 e −3kt .
Follow-up Question If k = 1, T1(0) = 10, T2(0) = 30,
Solution The initial condition for dT1
dt (0) is obtained with the
use of equation (I) above:
dt (0) = − 2k T1(0) + k T2(0) = − 20 + 30 = 10.
Thus, evaluating the given solution form and its derivative at
t = 0,
c1 e −0 + c2 e −3·0 = 10; − c1 e −0 − 3 c2 e −3·0 = 10.
From this we obtain c2 = − 10; c1 = 20 and the solution is
seen to be
T1(t) = 20 e −t − 10 e −3t ,
dt (t) = 30 e−3t − 20 e −2t .
Because the second block is relatively warm, the first block initially
warms up before ultimately cooling toward the temperature
of the table.