Solve an ODE using Fourier series

I've assumed ##y(t)## to be the sum of a complex Fourier series, and we get $$\sum (-n^2)c_ne^{int}+\sum ac_ne^{int}=\sum c_ne^{int}e^{in\pi},$$ which we can write as $$\sum ((-n^2)+a)c_ne^{int}=\sum (-1)^n c_ne^{int}.$$ We see here that equality holds if ##a=(-1)^n+n^2##. But how do I solve ##y''(t)+ay(t)=y(t+\pi)## when ##a=(-1)^n+n^2##. I don't think I understand the problem fully.

The functions ##e^{int}## are linearly independent, so you must have
$$\sum_{n=-\infty}^\infty \underbrace{[a-(n^2+(-1)^n)]c_n}_0 e^{int} = 0.$$ To get a non-trivial solution, ##a=m^2+(-1)^m## for some ##m \in \mathbb{Z}##. (Don't use ##n## because that's the dummy variable.)

What do you get if ##m=1##? ##m=2##?

vela said:

What do you get if ##m=1##? ##m=2##?

Hmm, we need ##[a-(n^2+(-1)^n)]c_n## to be zero for all ##n## and we want a non-trivial solution, so ##c_n## can’t be zero for all ##n##. It can be non-zero when ##n=m##. I guess we can allow for the coefficient ##n=-m## also to be non-zero, since ##a(m)## is even.

For example, if ##m=1##, we should get ##c_1e^{it}+c_{-1}e^{-it}## being the only terms that remain. The solution takes the form ##c_me^{imt}+c_{-m}e^{-imt}## for ##m\neq 0##. For ##m=0##, we get that the only coefficient that can be nonzero is ##c_0##, a constant solution.