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

1164 Chapter 17 Second-Order Differential Equations
Thus 23A − 1 and 2A 2 3B − 0, so A − 2 1 3 , B − 22 9 , and
For the equation y0 2 4y − cos 2x, we try
y p1 sxd − (2 1 3 x 2 2 9)e x
In Figure 3 we show the particular
solution y p − y p1 1 y p 2
of the differential
equation in Example 4. The
other solutions are given in terms of
f sxd − e 2x and tsxd − e 22x .
y p +f
_4 1
y p
FIGURE 3
y p +2f+g
y p +g
5
_2
y p2 sxd − C cos 2x 1 D sin 2x
Substitution gives
24C cos 2x 2 4D sin 2x 2 4sC cos 2x 1 D sin 2xd − cos 2x
or
28C cos 2x 2 8D sin 2x − cos 2x
Therefore 28C − 1, 28D − 0, and
y p2 sxd − 2 1 8 cos 2x
By the superposition principle, the general solution is
y − y c 1 y p1 1 y p2 − c 1 e 2x 1 c 2 e 22x 2 ( 1 3 x 1 2 9 )e x 2 1 8 cos 2x
■
Finally we note that the recommended trial solution y p sometimes turns out to be a
solution of the complementary equation and therefore can’t be a solution of the nonhomogeneous
equation. In such cases we multiply the recommended trial solution by x (or by
x 2 if necessary) so that no term in y p sxd is a solution of the complementary equation.
ExamplE 5 Solve y0 1 y − sin x.
SOLUTION The auxiliary equation is r 2 1 1 − 0 with roots 6i, so the solution of the
complementary equation is
Ordinarily, we would use the trial solution
y c sxd − c 1 cos x 1 c 2 sin x
y p sxd − A cos x 1 B sin x
but we observe that it is a solution of the complementary equation, so instead we try
y p sxd − Ax cos x 1 Bx sin x
Then
y p 9sxd − A cos x 2 Ax sin x 1 B sin x 1 Bx cos x
y p 0sxd − 22A sin x 2 Ax cos x 1 2B cos x 2 Bx sin x
Substitution in the differential equation gives
y p 0 1 y p − 22A sin x 1 2B cos x − sin x
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