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

1162 Chapter 17 Second-Order Differential Equations
Then y p 9 − 2Ax 1 B and y p 0 − 2A so, substituting into the given differential equation,
we have
s2Ad 1 s2Ax 1 Bd 2 2sAx 2 1 Bx 1 Cd − x 2
Figure 1 shows four solutions of the
differential equation in Example 1
in terms of the particular solution
y p and the functions f sxd − e x and
tsxd − e 22x .
y p +3g
8
y p +2f+3g
y p +2f
_3 3
y p
or 22Ax 2 1 s2A 2 2Bdx 1 s2A 1 B 2 2Cd − x 2
Polynomials are equal when their coefficients are equal. Thus
22A − 1 2A 2 2B − 0 2A 1 B 2 2C − 0
The solution of this system of equations is
A − 2 1 2 B − 21 2 C − 23 4
A particular solution is therefore
y p sxd − 2 1 2 x 2 2 1 2 x 2 3 4
FIGURE 1
_5
and, by Theorem 3, the general solution is
y − y c 1 y p − c 1 e x 1 c 2 e 22x 2 1 2 x 2 2 1 2 x 2 3 4
■
If Gsxd (the right side of Equation 1) is of the form Ce k x , where C and k are constants,
then we take as a trial solution a function of the same form, y p sxd − Ae k x , because the
derivatives of e k x are constant multiples of e k x .
Figure 2 shows solutions of the differential
equation in Example 2
in terms of y p and the functions
f sxd − cos 2x and tsxd − sin 2x.
Notice that all solutions approach ` as
x l ` and all solutions (except y p)
resemble sine func tions when x is
negative.
y p +f+g
y p +g
y p
_4 2
y p +f
FIGURE 2
4
_2
Example 2 Solve y0 1 4y − e 3x .
SOLUtion The auxiliary equation is r 2 1 4 − 0 with roots 62i, so the solution of the
complementary equation is
y c sxd − c 1 cos 2x 1 c 2 sin 2x
For a particular solution we try y p sxd − Ae 3x . Then y p 9 − 3Ae 3x and y p 0 − 9Ae 3x . Substituting
into the differential equation, we have
9Ae 3x 1 4sAe 3x d − e 3x
so 13Ae 3x − e 3x and A − 1
13 . Thus a particular solution is
and the general solution is
y p sxd − 1
13 e 3x
ysxd − c 1 cos 2x 1 c 2 sin 2x 1 1
13 e 3x ■
If Gsxd is either C cos kx or C sin kx, then, because of the rules for differentiating the
sine and cosine functions, we take as a trial particular solution a function of the form
y p sxd − A cos kx 1 B sin kx
Example 3 Solve y0 1 y9 2 2y − sin x.
SOLUtion We try a particular solution
y p sxd − A cos x 1 B sin x
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