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