Hey all,

heres the differential equation to be solved using Variation of parameter's:

y" + y = sin(x)^2

i got this far:

u1'cos(2x) + u2'sin(2x) =0

-2*u1'*sin(2x) + 2*u2'*cos(2x) = sin(x)^2

solving for u1' gives: (-sin(x)^2*sin(2x))/2

and u2': (-sin(x)^2*cos(2x))/2

The final solution should take the form

u1*cos(2x) + u2*sin(2x)

so whats left is integrating u1' and u2'. Unfortunately due to problems with integrating and simplfying trig expressions i'm not getting the solution which is

1/8(1-xsin(2x))

thanks alot