James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 17.2 Nonhomogeneous Linear Equations 1165
The graphs of four solutions of the
differential equation in Example 5 are
shown in Figure 4.
4
so A − 2 1 2 , B − 0, and
y
The general solution is
p sxd − 2 1 2 x cos x
ysxd − c 1 cos x 1 c 2 sin x 2 1 2 x cos x
■
_2π
y p
2π
We summarize the method of undetermined coefficients as follows:
FIGURE 4
_4
Summary of the Method of Undetermined Coefficients
1. If Gsxd − e kx Psxd, where P is a polynomial of degree n, then try
y p sxd − e kx Qsxd, where Qsxd is an nth-degree polynomial (whose coefficients
are determined by substituting in the differential equation).
2. If Gsxd − e kx Psxd cos mx or Gsxd − e kx Psxd sin mx, where P is an nth-degree
polynomial, then try
y p sxd − e kx Qsxd cos mx 1 e kx Rsxd sin mx
where Q and R are nth-degree polynomials.
Modification: If any term of y p is a solution of the complementary equation,
multiply y p by x (or by x 2 if necessary).
ExamplE 6 Determine the form of the trial solution for the differential equation
y0 2 4y9 1 13y − e 2x cos 3x.
SOLUTION Here Gsxd has the form of part 2 of the summary, where k − 2, m − 3, and
Psxd − 1. So, at first glance, the form of the trial solution would be
y p sxd − e 2x sA cos 3x 1 B sin 3xd
But the auxiliary equation is r 2 2 4r 1 13 − 0, with roots r − 2 6 3i, so the solution
of the complementary equation is
y c sxd − e 2x sc 1 cos 3x 1 c 2 sin 3xd
This means that we have to multiply the suggested trial solution by x. So, instead, we
use
y p sxd − xe 2x sA cos 3x 1 B sin 3xd
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The Method of Variation of Parameters
Suppose we have already solved the homogeneous equation ay0 1 by9 1 cy − 0 and
written the solution as
4 ysxd − c 1 y 1 sxd 1 c 2 y 2 sxd
where y 1 and y 2 are linearly independent solutions. Let’s replace the constants (or parameters)
c 1 and c 2 in Equation 4 by arbitrary functions u 1 sxd and u 2 sxd. We look for a particu-
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