Solving $f''+f=\sin x$ with Fourier series Does there exist a twice differentiable periodic function $f$ such that
$f''(x) + f(x) =\sin(x)$ for all $x \in [-\pi, \pi]$?
How to solve this differential equation using Fourier series? I know only basics of Fourier analysis. I don`t know any inversion formula for Fourier series.
 A: The solution would be:
$$f(x) = -\frac{1}{2}x \cos(x) + C_1\sin(x) + C_2\cos(x)$$
Where the $C_1\sin(x) + C_2\cos(x)$ part is the solution to the homogeneous equation. Using Fourier Series naively, one runs into problems due to the $n = 1$ term having no solution. 
A: I'm lazy and don't feel like giving you the complete answer, but here is an approach you can take. 
Take the Fourier transform of both sides:
$F'' + F' = T[\sin(x)]$
Fourier transforms differentiate easily:
$$F' = ik\cdot F$$ and 
$$F'' = -k^2\cdot F$$
So, then solve for $F$:
$$F = \frac{T[\sin(x)]} {ik-k^2}$$
Then, take the inverse transform:
$$f = T^{-1}\left[\frac{T[\sin(x)]}{ik-k^2}\right]$$
You can solve the right-hand side either by looking up the transforms, or using integration by parts.
A: nbubis mentions that Fourier Series might not be the best method.
Suppose that $f$ has a Fourier Series:
$$
f(x)=\sum_{k=-\infty}^\infty a_ke^{ikx}\tag{1}
$$
Then
$$
f''(x)+f(x)=\sin(x)\tag{2}
$$
implies
$$
\sum_{k=-\infty}^\infty a_k(1-k^2)e^{ikx}=\sin(x)=\frac{1}{2i}\left(e^{ix}-e^{-ix}\right)\tag{3}
$$
However, integrating $(3)$ against $\frac{1}{2\pi}e^{-ikx}$ to get $a_k$, gives $a_k=0$ except for $k\in\{-1,1\}$. Equation $(3)$ says that $a_{-1}\cdot0=-\frac{1}{2i}$ and $a_1\cdot0=\frac{1}{2i}$. This leads one to conclude that there is no solution to $(2)$ which has a Fourier series.
