Hello,

The problem I am trying to solve is the following: I am to find the steady periodic solution of the differential equation.

F(t)= x''+2x where F is the even function of period 2pi such that F = sin(t) if 0<t<pi. I assume xp has the form A_{n}Cos(nt)+B_{n}Sin(nt), B =0 here

I did a fourier series expansion for sin (t) and got 2/pi +((2/pi) * summation from n=1 (n even) to infinity of (2/(1-N^{2})). I now went on to equate coefficients to find a xp with a known value of A, but got the incorrect answer. Is my procedure so far correct?