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2.3 The Nonhomogeneous Equation 55

2.3 The Nonhomogeneous Equation

From Theorem 2.4, the keys to solving the nonhomogeneous linear equation (2.1) are to find two
linearly independent solutions of the associated homogeneous equation and a particular solution
Y p for the nonhomogeneous equation. We can perform the first task at least when the coefficients
are constant. We will now focus on finding Y p , considering two methods for doing this.

2.3.1 Variation of Parameters
Suppose we know two linearly independent solutions y 1 and y 2 of the associated homogeneous
equation. One strategy for finding Y p is called the method of variation of parameters. Look for
functions u 1 and u 2 so that
Y p (x) = u 1 (x)y 1 (x) + u 2 (x)y 2 (x).
To see how to choose u 1 and u 2 , substitute Y p into the differential equation. We must compute
two derivatives. First,
Y = u 1 y + u 2 y + u y 1 + u y 2 .

1
p
2
2
1
Simplify this derivative by imposing the condition that

u y 1 + u y 2 = 0. (2.8)
1 2
Now

Y = u 1 y + u 2 y ,

p 1 2
so

Y = u y + u y + u 1 y + u 2 y .
p 1 1 2 2 1 2
Substitute Y p into the differential equation to get
u y + u y + u 1 y + u 2 y

1 1 2 2 1 2
+ p(x)(u 1 y + u 2 y ) + q(x)(u 1 y 1 + u 2 y 2 ) = f (x).

1 2
Rearrange terms to write

u 1 [y + p(x)y + q(x)y 1 ]
1
1

+ u 2 [y + p(x)y + q(x)y 2 ]
2 2

+ u y + u y = f (x).
1 1 2 2

The two terms in square brackets are zero, because y 1 and y 2 are solutions of y + p(x)y +
q(x)y = 0. The last equation therefore reduces to

u y + u y = f (x). (2.9)

1 1 2 2
Equations (2.8) and (2.9) can be solved for u and u to get

2
1
y 2 (x) f (x) y 1 (x) f (x)

u (x) =− and u (x) = (2.10)
1 2
W(x) W(x)
where W(x) is the Wronskian of y 1 (x) and y 2 (x).
We know that W(x) =0, because y 1 and y 2 are assumed to be linearly independent solutions
of the associated homogeneous equation. Integrate equations (2.10) to obtain
y 2 (x) f (x) y 1 (x) f (x)

u 1 (x) =− dx and u 2 (x) = dx. (2.11)
W(x) W(x)
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October 14, 2010 14:12 THM/NEIL Page-55 27410_02_ch02_p43-76