Define $PC^r(2\pi)=\{f:[-\pi,\pi]\rightarrow\mathbb{R}: f\in\mathcal{C}^r \text{ and } f,f',f'',\dots, f^{(r)} \text{ are } 2\pi\text{-periodic} \}$. I want to show that if $g\in PC^1(2\pi)$ and $f\in PC^2(2\pi)$ and $f''(x) + λf(x) = g(x)$ such that $\lambda \neq n^2$ for all $n\in\mathbb{N}$, then I can recover the coefficients of the fourier expansion of $g$ as a function of the coefficients of the function of $f$. This seems very intuitive, but I don't seem to be able to show it. If anyone has a fair hint I would be very appreciative.
Update: So I believe I have managed to come up with a solution that I want to expose so that you guys argue whether it is correct or not.
First of all we solve for the homogeneous part to the solution of $f$, so we get from the characteristic equation that $$f_h(x) = A \cos (\sqrt{\lambda}t) + B\sin(\sqrt{\lambda}t)$$
We now solve the particular part of the solution to $f$.We know that as $g\in PC^1(2\pi)$ then it should have a Fourier expansion so that $$g(t) = a_0 + \sum_{n\in\mathbb{N}} \left[a_n \cos(nt) + b_n \sin(nt)\right]$$ so we can propose a solution of the form $$f_p(t) = \alpha_0 + \sum_{n\in\mathbb{N}} \left[\alpha_n \cos(nt) + \beta_n \sin(nt)\right]$$ for which we will solve via undetermined coefficients, so we differentiate twice and substitute on the differential equation to get that
$$\lambda\alpha_0 + \sum_{n\in\mathbb{N}} \left[(\lambda-n^2 )\alpha_n \cos(nt) +(\lambda-n^2 ) \beta_n \sin(nt)\right] = a_0 + \sum_{n\in\mathbb{N}} \left[a_n \cos(nt) + b_n \sin(nt)\right]$$ and with this equation we get $\alpha_0 = a_0/\lambda$, $\alpha_n = a_n/(\lambda-n^2)$ and $\beta_n = b_n/(\lambda-n^2)$. Therefore the solution to the equation is $f(x) = f_h(x) + f_p(x)$. Now as we have, by assumption that $f\in PC^2(2\pi)$ then $f(0) = f(2\pi)$ and $f(\pi) = f(-\pi)$ which implies that we necessarily need $A=B=0$ from the homogeneous solution as, by assumption $\lambda \neq n^2$ for any $n\in\mathbb{N}$.
With this, we have managed to recover $a_0, a_n, b_n$ for all $n$ as mentioned above.