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## Homework Statement

Find a particular solution of the given differential equation.

y'' + 4y = (sin^2)(x)

Answer from book: yp(x)= (1/8)(1-xsin2x)

## The Attempt at a Solution

y'' + 4y = 4(sinx)^2....(1)

4(sinx)^2=2 - 2cos2x

Homogeneous solution:

r^2+4=0

r^2= -4

r=+-2i ==>

yh=C1cos(2x)+C2sin(2x)

Particular solution:

yp=A+Bxcos2x+Cxsin2x

yp' =Bcos2x-2Bxsin2x+Csin2x+2Cxcos2x

yp" = -2Bsin2x -2Bsin2x-4Bxcos2x

+2Ccos2x+2Ccos2x - 4Cxsin2x

Substitute y and y" in (1) ==>

4A - 4Bsin2x+4Ccos2x=2 - 2cos2x ==>

A=1/2

B=0

C= -1/2

yp=1/2 - (1/2)xsin2x

This is answer is different from the book. What am I doing wrong?