Solve an ODE using Fourier series

[psie]

Homework Statement

Find the values of the constant $a$ for which the problem $y''(t)+ay(t)=y(t+\pi), \ t\in\mathbb R$, has a solution with period $2\pi$ which is not identically zero. Determine all such solutions.

Relevant Equations

The complex Fourier series of a $2\pi$ periodic function, namely $\sum_{n\in\mathbb Z} c_ne^{int}$.

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.

 

[vela]

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 $m=1$? $m=2$?

 

[psie]

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.