MATH 2280-002
Lecture Notes: 2/08/2013
Math 2280 Section 002 [SPRING 2013]
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Variation of Parameters
Today we’ll learn about the second method for solving nonhomogeneous, linear DE’s.
Variation of Parameters. If y1 , y2 ,..., yn are n linearly independent solutions to the homogeneous
DE
p0 (x)y (n) + p1 (x)y (n−1) + ... + pn−1 (x)y 0 + pn (x)y = 0,
then a particular solution to
p0 (x)y (n) + p1 (x)y (n−1) + ... + pn−1 (x)y 0 + pn (x)y = f (x),
is
yp = c1 y1 + c2 y2 + ... + cn yn ,
Z
where ci =
Wi (y1 , y2 , ..., yn )
dx.
W (y1 , y2 , ..., yn )
Here W (y1 , y2 , ..., yn ) is the Wronskian, and Wi (y1 , y2 , ..., yn ) is the determinant of the same
matrix as the Wronskian, except that the ith column is replaced by
(0, 0, ..., 0, f (x)).
Let’s consider the n = 2 case. (This is the only case I’ll expect you to know for this class.) Suppose
y1 and y2 are linearly independent solutions to
p0 (x)y 00 + p1 (x)y 0 + p2 (x)y = 0.
y1 y2 = y1 y 0 − y2 y 0
• W (y1 , y2 ) = 0
2
1
y1 y20 0
y2 • W1 (y1 , y2 ) = = −y2 f (x)
f (x) y20 y1
0 = y1 f (x)
• W2 (y1 , y2 ) = 0
y1 f (x) That means a particular solution to
p0 (x)y 00 + p1 (x)y 0 + p2 (x)y = f (x)
is
Z
yp (x) = y1
−y2 f (x)
dx + y2
y1 y20 − y2 y10
Z
y1 f (x)
dx.
y1 y20 − y2 y10
Yeah, I know that’s ugly, but it works even when the method of undetermined coefficients doesn’t
apply. Still, I would recommend using undetermined coefficients whenever you can.
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MATH 2280-002
Lecture Notes: 2/08/2013
Example. Find the general solution for
y 00 − 9y = e3x .
This a DE we can solve using undetermined coefficients, but let’s use variation of parameters to
compare the methods. Last time we found that, in this case,
yh (x) = c1 e3x + c2 e−3x .
That means y1 (x) = e3x
3x
e
• W (y1 , y2 ) = 3x
3e
0
• W1 (y1 , y2 ) = 3x
e
3x
e
• W2 (y1 , y2 ) = 3x
3e
and y2 (x) = e−3x .
e−3x = −6
−3e−3x e−3x = −1
−3e3x 0 = e6x
e3x A particular solution is given by
Z 6x
e
−1
dx + e−3x
dx
−6
−6
6x
e
3x 1
−3x
−3x
=e
x + c1 + e
+ c2 .
+e
−
6
36
yp (x) = e3x
Z
Every particular solution has to be of this form for some choice of c1 and c2 . That means we can
make a choice for c1 and c2 (for example, c1 = c2 = 0) to get a particular solution, or we could
just leave c1 and c2 as unknown constants and obtain the general solution:
1 3x
e3x
1 3x e3x
3x
−3x
+ c1 e + c2 e
= xe + c1 −
e3x + c2 e−3x .
y(x) = xe −
6
36
6
36
If you work this out using undetermined coefficients, you will get a different particular solution. In
fact, you get
1
y(x) = xe3x + d1 e3x + d2 e−3x .
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But that’s okay. Do you see why?
Exercise. Find the general solution for
y 00 + y = sec(x).
This is a DE we don’t know how to solve using the method of undetermined coefficients, so variation
of parameters is our only choice, We’ll work this out in class, but try to solve it on your own.
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