# Differential Equation NonHomogeneous EQ

## 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?

## Answers and Replies

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Mark44
Mentor

## 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.