The eqn is :

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

and i'm in need of a particular solution, y(p).

I tried y= B/2 - A/2cos(2x)(below)

y = B/2 - A/2cos(2x)

y' = Asin(2x)

y'' = 2Acos(2x)

Next,

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

2Acos(2x)+Asin(2x)+ B/2 - A/2cos(2x) = (sin(x))^2

I can't anywhere with this.

The given answer is y(p) = 1/26(13+3cos(2x) -2sin(2x))

best,

jblorien