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

Section 17.2 Nonhomogeneous Linear Equations 1163
Then y p 9 − 2A sin x 1 B cos x y p 0 − 2A cos x 2 B sin x
so substitution in the differential equation gives
s2A cos x 2 B sin xd 1 s2A sin x 1 B cos xd 2 2sA cos x 1 B sin xd − sin x
or
s23A 1 Bd cos x 1 s2A 2 3Bd sin x − sin x
This is true if
The solution of this system is
so a particular solution is
23A 1 B − 0 and 2A 2 3B − 1
A − 2 1
10 B − 2 3
10
y p sxd − 2 1
10 cos x 2 3
10 sin x
In Example 1 we determined that the solution of the complementary equation is
y c − c 1 e x 1 c 2 e 22x . Thus the general solution of the given equation is
ysxd − c 1 e x 1 c 2 e 22x 2 1
10 scos x 1 3 sin xd ■
If Gsxd is a product of functions of the preceding types, then we take the trial solution
to be a product of functions of the same type. For instance, in solving the differential
equation
we would try
y0 1 2y9 1 4y − x cos 3x
y p sxd − sAx 1 Bd cos 3x 1 sCx 1 Dd sin 3x
If Gsxd is a sum of functions of these types, we use the easily verified principle of
superposition, which says that if y p1 and y p2 are solutions of
ay0 1 by9 1 cy − G 1 sxd
ay0 1 by9 1 cy − G 2 sxd
respectively, then y p1 1 y p2 is a solution of
ay0 1 by9 1 cy − G 1 sxd 1 G 2 sxd
Example 4 Solve y0 2 4y − xe x 1 cos 2x.
SOLUtion The auxiliary equation is r 2 2 4 − 0 with roots 62, so the solution of the
complementary equation is y c sxd − c 1 e 2x 1 c 2 e 22x . For the equation y0 2 4y − xe x
we try
y p1 sxd − sAx 1 Bde x
Then y p1 9 − sAx 1 A 1 Bde x , y p1 0 − sAx 1 2A 1 Bde x , so substitution in the equation
gives
sAx 1 2A 1 Bde x 2 4sAx 1 Bde x − xe x
or
s23Ax 1 2A 2 3Bde x − xe x
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