MA 222
Proof of the Principle of Superposition
K. Rotz
Theorem (known as the Principle of Superposition): Consider the second-order, linear, homogeneous
ordinary differential equation
p(x)y 00 (x) + q(x)y 0 (x) + r(x)y(x) = 0.
(∗)
If y1 and y2 are both solutions to (∗), then for any two constants c1 and c2 ,
y = c1 y1 + c2 y2
is also a solution to (∗).
Proof : The fact that y1 and y2 are solutions to (∗) imply that
p(x)y100 (x) + q(x)y10 (x) + r(x)y1 (x) = 0
p(x)y200 (x) + q(x)y20 (x) + r(x)y2 (x) = 0.
and
(1)
(2)
Since c1 and c2 are constants, we have
y 0 (x) = c1 y10 (x) + c2 y20 (x),
y 00 (x) = c1 y100 (x) + c2 y20 (x).
and
Inserting these into (∗), we see that
y 0 (x)
y 00 (x)
}|
{
z
}|
{
z
p(x)y 00 (x) + q(x)y 0 (x) + r(x)y(x) = p(x)(c1 y100 (x) + c2 y20 (x)) + q(x)(c1 y10 (x) + c2 y20 (x))
+ r(x)(c1 y1 (x) + c2 y2 (x))
|
{z
}
(3)
y(x)
We now regroup the terms in (3) by those terms with c1 ’s and c2 ’s:
p(x)y 00 (x) + q(x)y 0 (x) + r(x)y(x) = c1 p(x)y100 (x) + c2 p(x)y20 (x) + c1 q(x)y10 (x) + c2 q(x)y20 (x)
+ c1 r(x)y1 (x) + c2 r(x)y2 (x)
= c1 [p(x)y100 (x) + q(x)y10 (x) + r(x)y1 (x)]
+ c2 [p(x)y200 (x) + q(x)y20 (x) + r(x)y2 (x)]
By equations (1) and (2), the right-hand side of (4) is zero. In other words,
p(x)y 00 (x) + q(x)y 0 (x) + r(x)y(x) = 0,
so that y is a solution to (∗).
1
(4)