Differential Equation NonHomogeneous EQ

[JJBrian]

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?

 

[Mark44]

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)

How did the right side go from sin²(x) to 4sin²(x)?

What you need to do is rewrite the right side using a double angle identity, sin²(x) = 1/2 - (1/2)cos(2x).
Your work below suggests that you have done this, but I don't see that you mention it anywhere.

4(sinx)^2=2 - 2cos2x

Homogeneous solution:
r^2+4=0
r^2= -4
r=+-2i ==>
yh=C1cos(2x)+C2sin(2x)

This is fine.

Particular solution:
yp=A+Bxcos2x+Cxsin2x

This (above) is what you need.

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?

The only mistake I see is that you changed the right side of the DE from sin²(x) to 4sin²(x). That could be causing problems for you in solving for the coefficients A, B, and C.