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

Section 17.2 Nonhomogeneous Linear Equations 1167
Then
y p 0 − u 1 9 cos x 2 u 2 9 sin x 2 u 1 sin x 2 u 2 cos x
For y p to be a solution we must have
11 y p 0 1 y p − u 1 9 cos x 2 u 2 9 sin x − tan x
Solving Equations 10 and 11, we get
u 1 9ssin 2 x 1 cos 2 xd − cos x tan x
u 1 9 − sin x
u 1 sxd − 2cos x
(We seek a particular solution, so we don’t need a constant of integration here.) Then,
from Equation 10, we obtain
Figure 5 shows four solutions of the
differential equation in Example 7.
2.5
So
u 2 9 − 2 sin x
cos x u 19 − 2 sin2 x
cos x − cos2 x 2 1
− cos x 2 sec x
cos x
u 2 sxd − sin x 2 lnssec x 1 tan xd
(Note that sec x 1 tan x . 0 for 0 , x , y2.) Therefore
y p sxd − 2cos x sin x 1 fsin x 2 lnssec x 1 tan xdg cos x
0
y p
_1
π
2
− 2cos x lnssec x 1 tan xd
and the general solution is
FIGURE 5
ysxd − c 1 sin x 1 c 2 cos x 2 cos x lnssec x 1 tan xd
■
1–10 Solve the differential equation or initial-value problem
using the method of undetermined coefficients.
1. y0 1 2y9 2 8y − 1 2 2x 2
2. y0 2 3y9 − sin 2x
3. 9y0 1 y − e 2x
4. y0 2 2y9 1 2y − x 1 e x
5. y0 2 4y9 1 5y − e 2x
6. y0 2 4y9 1 4y − x 2 sin x
7. y0 2 2y9 1 5y − sin x, ys0d − 1, y9s0d − 1
8. y0 2 y − xe 2x , ys0d − 0, y9s0d − 1
9. y0 2 y9 − xe x , ys0d − 2, y9s0d − 1
10. y0 1 y9 2 2y − x 1 sin 2x, ys0d − 1, y9s0d − 0
;
11–12 Graph the particular solution and several other solutions.
What characteristics do these solutions have in common?
11. y0 1 3y9 1 2y − cos x 12. y0 1 4y − e 2x
13–18 Write a trial solution for the method of undetermined
coefficients. Do not determine the coefficients.
13. y0 2 y9 2 2y − xe x cos x
14. y0 1 4y − cos 4x 1 cos 2x
15. y0 2 3y9 1 2y − e x 1 sin x
16. y0 1 3y9 2 4y − sx 3 1 xde x
17. y0 1 2y9 1 10y − x 2 e 2x cos 3x
18. y0 1 4y − e 3x 1 x sin 2x
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