I have that
$$f(x)=\frac{4}{3}+\frac{2}{\pi^2}\sum_{n\in\mathbb{Z}}\frac{(-1)^n(1+i\pi n)}{n^2}e^{i\pi n x}, \quad n\neq0\tag1$$
and I want to find a 2-periodic function $y(x)$ that solves the following diff-equation
$$2y''-y'-y=f(x).\tag{2}$$
Since $(1)$ is a linear ODE I know that it's solution is of the form $y(x)=y_h+y_p.$ Finding $y_h$ is trivial and done by solving the characteristic equation. I found it to be $y_h=C_1e^x+C_2e^{-x/2}.$
For $y_p$, I need to use $f(x)$ somehow, I could not find any good example in the book on how to do this. I only found one video on youtube but it does not really seem to be that similar to my problem.
I want to follow the answer in this thread but I can't seem to grasp the mechanics of what he means. Can someone show me how to go about this?
A $2$-periodic function has the form
$$ y(x) = c_0 + \sum_{n=\pm 1}^{\pm\infty} c_n e^{in\pi x} $$
where $c_n$ are unknown coefficients. Plug this expression into the ODE (differentiating term-wise)
$$ 2y'' - y' - y = - c_0 - \sum_{n=\pm1}^{\pm\infty} (n^2\pi^2 + in\pi + 1)c_n e^{in\pi x} $$
Comparing coefficients gives
\begin{align} -c_0 &= \frac43 \\ -(n^2\pi^2+in\pi+1)c_n &= \frac{2}{\pi^2}(-1)^n \frac{1+in\pi}{n^2} \end{align}
The homogeneous solution is not periodic so it isn't needed.