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

1166 Chapter 17 Second-Order Differential Equations
lar solution of the nonhomogeneous equation ay0 1 by9 1 cy − Gsxd of the form
5 y p sxd − u 1 sxd y 1 sxd 1 u 2 sxd y 2 sxd
(This method is called variation of parameters because we have varied the parameters
c 1 and c 2 to make them functions.) Differentiating Equation 5, we get
6 y p 9 − su 1 9y 1 1 u 2 9y 2 d 1 su 1 y 1 9 1 u 2 y 2 9d
Since u 1 and u 2 are arbitrary functions, we can impose two conditions on them. One
con dition is that y p is a solution of the differential equation; we can choose the other
condition so as to simplify our calculations. In view of the expression in Equation 6, let’s
impose the condition that
7 u 1 9y 1 1 u 2 9y 2 − 0
Then y p 0 − u 1 9y 1 9 1 u 2 9y 2 9 1 u 1 y 1 0 1 u 2 y 2 0
Substituting in the differential equation, we get
or
asu 1 9y 1 9 1 u 2 9y 2 9 1 u 1 y 1 0 1 u 2 y 2 0d 1 bsu 1 y 1 9 1 u 2 y 2 9d 1 csu 1 y 1 1 u 2 y 2 d − G
8 u 1 say 1 0 1 by 1 9 1 cy 1 d 1 u 2 say 2 0 1 by 2 9 1 cy 2 d 1 asu 1 9y 1 9 1 u 2 9y 2 9d − G
But y 1 and y 2 are solutions of the complementary equation, so
and Equation 8 simplifies to
ay 1 0 1 by 1 9 1 cy 1 − 0 and ay 2 0 1 by 2 9 1 cy 2 − 0
9 asu 1 9y 1 9 1 u 2 9y 2 9d − G
Equations 7 and 9 form a system of two equations in the unknown functions u 1 9 and u 2 9.
After solving this system we may be able to integrate to find u 1 and u 2 and then the particular
solution is given by Equation 5.
ExamplE 7 Solve the equation y0 1 y − tan x, 0 , x , y2.
SOLUTION The auxiliary equation is r 2 1 1 − 0 with roots 6i, so the solution of
y0 1 y − 0 is ysxd − c 1 sin x 1 c 2 cos x. Using variation of parameters, we seek a
solution of the form
y p sxd − u 1 sxd sin x 1 u 2 sxd cos x
Then
y p 9 − su 1 9 sin x 1 u 2 9 cos xd 1 su 1 cos x 2 u 2 sin xd
Set
10 u 1 9 sin x 1 u 2 9 cos x − 0
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