REDUCTION OF THE ORDER
MINGFENG ZHAO
September 22, 2014
Reduction of order method
Consider the homogeneous equation:
y 00 + p(x)y 0 + q(x)y = 0.
Question 1. If we have already known a solution y1 to y 00 + p(x)y 0 + q(x)y = 0, how can we find another solution y2
such that y1 and y2 are linearly independent?
Assume y2 is another solution to y 00 + p(x)y 0 + q(x)y = 0, let
v(x) =
y2 (x)
.
y1 (x)
That is, y2 (x) = y1 (x)v(x). We are going to find what v(x) is. Since y2 is a solution to y 00 + p(x)y 0 + q(x)y = 0, then
0
= y200 + p(x)y20 + q(x)y2
=
[y1 (x)v(x)]00 + p(x)[y1 (x)v(x)]0 + q(x)[y1 (x)v(x)]
=
[v(x)y10 + y1 v 0 (x)]0 + p(x)[v(x)y10 + y1 v 0 (x)] + v(x)q(x)y1 (x)
= v(x)y100 + v 0 (x)y10 + y10 v 0 (x) + y1 v 00 (x) + v(x)p(x)y10 + v 0 (x)p(x)y1 + v(x)q(x)y1 (x)
= v(x)[y100 + p(x)y10 + q(x)y1 ] + 2v 0 (x)y10 + y1 (x)v 00 (x) + p(x)y1 (x)v 0 (x)
=
2v 0 (x)y10 (x) + y1 (x)v 00 (x) + p(x)y1 (x)v 0 (x),
Since y1 is a solution to y 00 + p(x)y 0 + q(x)y = 0
So we get a differential equation of v:
y1 (x)v 00 + [2y10 (x) + p(x)y1 (x)]v 0 = 0.
That is,
2y 0 (x) + p(x)y1 (x) 0
·v =−
v =− 1
y1 (x)
00
1
2y10 (x)
+ p(x) · v 0 .
y1 (x)
2
MINGFENG ZHAO
Let w(x) = v 0 (x), then
w0 = −
2y10 (x) + p(x)y1 (x)
· w,
y1 (x)
which is a separable equation. Then
−
w(x) = e
R
0 (x)
2y1
+p(x)
y1 (x)
dx
R
=e
R
−2 ln y1 (x)− p(x) dx
e− p(x) dx
=
.
[y1 (x)]2
So we get
R
e− p(x) dx
v (x) = w(x) =
.
[y1 (x)]2
0
Hence
Z
v(x) =
Z
0
v (x) dx =
R
e− p(x) dx
dx.
[y1 (x)]2
Therefore, we find another solution y2 to y 00 + p(x)y 0 + q(x)y = 0:
Z
y2 (x) = y1 (x)v(x) = y1 (x) ·
R
e− p(x) dx
dx .
[y1 (x)]2
Example 1. Solve t2 y 00 + ty 0 − 4y = 0, t > 0. First to check y1 (t) = t2 is a solution to t2 y 00 + ty 0 − 4y = 0. Find another
solution y2 which is linearly independent to y1 .
In fact, since y1 (t) = t2 , then y10 (t) = 2t and y20 (t) = 2. So we have
t2 y100 + ty10 − 4y1 = 2t2 + 2t2 − 4t2 = 0.
Hence
y1 (t) = t2 is a solution to t2 y 00 + ty 0 − 4y = 0 .
Let y2 (t) = v(t)y1 (t) = t2 v be another solution to t2 y 00 + ty 0 − 4y = 0, then y20 (t) = 2tv + t2 v 0 and y200 (t) =
2v + 2tv 0 + 2tv 0 + t2 v 00 = t2 v 00 + 4tv 0 + 2v. So we have
0
= t2 y200 + ty20 − 4y2
= t2 [t2 v 00 + 4tv 0 + 2v] + t[2tv + t2 v 0 ] − 4t2 v
= t4 v 00 + 4t3 v 0 + 2t2 + 2t2 v + t3 v 0 − 4t2 v
= t4 v 00 + 5t3 v 0 .
That is, v satisfies:
5
v 00 = − v 0 .
t
REDUCTION OF THE ORDER
3
Let w(t) = v 0 (t), then w satisfies
5
w0 (t) = − · w(t).
t
So
w(t) = e−
R
5
t
dt
= e−5 ln t =
1
.
t5
Since v 0 (t) = w(t), then
Z
v(t) =
0
Z
1
4
dt = − 4 .
t5
t
v (t) dt =
So
y2 (t) = v(t)y1 (t) = −
4 2
4
· t = − 2.
t4
t
It’s obvious that y1 and y2 are linearly independent. So
y2 (t) = −
4
is another solution to t2 y 00 + ty 0 − 4y = 0 .
t2
Let y1 be a solution to y 00 +p(x)y 0 +q(x)y = 0, to find another solution y2 such that y1 and y2 are linearly independent:
1) Let y2 = v(x)y1 , plug y2 = v(x)y1 into y 00 + p(x)y 0 + q(x)y = 0. Since y1 is a solution to y 00 + p(x)y 0 + q(x)y = 0,
we will get That is,
2y 0 (x) + p(x)y1 (x) 0
·v =−
v =− 1
y1 (x)
00
2y10 (x)
+ p(x) · v 0 .
y1 (x)
2) Let w(x) = v 0 (x), then
w0 = −
2y10 (x) + p(x)y1 (x)
· w,
y1 (x)
3) Solve w, then
R
e− p(x) dx
.
v (x) = w(x) =
[y1 (x)]2
0
4) Solve v, then
Z
v(x) =
0
Z
v (x) dx =
R
e− p(x) dx
dx.
[y1 (x)]2
5) Write down the solution y2 to y 00 + p(x)y 0 + q(x)y = 0:
Z
y2 (x) = y1 (x)v(x) = y1 (x) ·
R
e− p(x) dx
dx .
[y1 (x)]2
4
MINGFENG ZHAO
1
1
Example 2. Find the general solution to u00 − u0 + 2 u = 0. An obvious solution is u1 (t) = t.
t
t
1
1
Let u2 (t) = v(t)u1 (t) = tv be another solution to u00 − u0 + 2 u = 0, then u02 (t) = v + tv 0 and u002 (t) = v 0 + v 0 + tv 00 =
t
t
tv 00 + 2v 0 . So we get
1
1
= tv 00 + 2v 0 − (v + tv 0 ) + 2 · tv
t
t
1
1
= tv 00 + 2v 0 − · v − v 0 + · v
t
t
0
= tv 00 + v 0 .
So v satisfies
1
v 00 = − · v 0 .
t
Let w(t) = v 0 , then
1
w0 = v 00 = − w.
t
So we have
w(t) = e
Since v 0 (t) = w(t) =
R
− 1t dt
1
, then
t
Z
v(t) =
= e− ln t =
1
.
t
1
dt = ln t.
t
So u2 (t) = tv(t) = t ln t. It’s easy to see that t ln t and t are linearly independent. Therefore, the general solution to
1
1
u00 − u0 + 2 u = 0 is:
t
t
u(t) = C1 t + C2 t ln t .
Example 3. Let a, b and c be three constants such that b2 = 4ac, find the general solution to ay 00 + by 0 + cy = 0. It’s
b
easy to see that y1 (x) = e− 2a ·x is a solution.
b
Let y2 (x) = y1 (x)v(x) = v(x)e− 2a ·x be anothe solution to ay 00 + by 0 + cy = 0, then
b
b
v(x)e− 2a ·x
2a
b
b
b
b
b
b
b2
y200 (x) = v 00 (x)e− 2a ·x − v 0 (x)e− 2a ·x − v 0 (x)e− 2a ·x + 2 v(x)e− 2a ·x
2a
2a
4a
b
b
b
b
b2
= v 00 (x)e− 2a ·x − v 0 (x)e− 2a ·x + 2 v(x)e− 2a ·x .
a
4a
y20 (x)
b
= v 0 (x)e− 2a ·x −
Since ay200 + by20 + cy2 = 0, then
0
b
b
= av 00 (x)e− 2a ·x − bv 0 (x)e− 2a ·x +
b
b2
v(x)e− 2a ·x
4a
REDUCTION OF THE ORDER
b
+bv 0 (x)e− 2a ·x −
5
b
b2
v(x)e− 2a ·x
2a
b
+cv(x)e− 2a ·x
b
b
b2
= av 00 (x)e− 2a ·x + c −
v(x)e− 2a ·x
4a
b
= av 00 (x)e− 2a
Since b2 = 4ac.
b
So we get v 00 (x) = 0, we can take v(x) = x, then y2 (x) = xe− 2a ·x is anoth solution to ay 00 + by 0 + cy = 0. Hence the
general solution to ay 00 + by 0 + cy = 0 is:
b
b
y(x) = C1 e− 2a ·x + C2 xe− 2a ·x .
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca