Reduction of Order
Suppose that y1 is a solution to the problem
y 00 + p(x)y 0 + q(x)y = 0
(1)
Our goal is to find a second linearly independent solution y2 .
The motivation for this approach is the method of variation of parameters seen earlier in the class. We
seek a solution in the form
y2 (x) = v(x)y1 (x).
This implies that
y20 = v 0 y1 + vy10 ,
and
y200 = v 00 y1 + 2v 0 y10 + vy100 .
Substituting these expressions into (1) gives
00
v y1 + 2v 0 y10 + vy100 + p(x) v 0 y1 + vy10 + q(x) vy1 = 0.
Collecting the terms multiplying v we get
v(y100 + p(x)y10 + q(x)y1 ) = 0
so the equation for v simplifies to
y1 v 00 + (2y10 + p(x)y1 )v 0 = 0.
Thus we obtain
v 00 +
Setting w = v 0 and P =
(2y10 + p(x)y1 ) 0
v = 0.
y1
(2)
(2y10 + p(x)y1 )
this equation reduces to the first order linear equation
y1
w0 + P w = 0.
Z
A solution is w = exp −
Z
v = exp −
0
x
P (s) ds so that
x
P (s) ds
Z
= exp −
and hence
Z
v=
So we have
Z
2y10
1
+ p(x) ds = 2 exp −
y1
y1
Z
1
exp −
y12 (x)
Z
y2 = y1
x
x
Z
1
exp −
y12 (x)
p(x) ds
x
dx.
p(x) ds dx.
x
p(x) ds
As an example consider y 00 − 2r0 y 0 + r02 y = 0. The characteristic equation has a double root r = r0 so
one solution is y1 = er0 x . Here p(x) = −2r0 and q(x) = r02 . We seek y2 = vy1 and obtain :
Z x
Z
Z
1
r0 x
r0 x
dx = xer0 x .
y2 = e
exp
−
−2r
ds
dx
=
e
0
e2r0 x