Find all the functions that hold the equation $e^{ix}f''(x)+5f'(x)+f(x)=0$

Question

Given $f:\mathbb{R}\rightarrow\mathbb{R},f\in C^{2}(\mathbb{T})$
find all the function that hold $e^{ix}f''(x)+5f'(x)+f(x)=0$.
I've used the Fourier coefficients and got:
$$\hat{f}(n)=\frac{(n-1)^{2}\hat{f}(n-1)}{1-5in}$$. for every $n\geq 1$ I've should that all the coefficients are $0$ and for the negative i didn't managed to show something.

Answer 1

The equation $\cos(x)f''+i\sin(x)f''+5f'+f=0$ is equivalent to the system

$$\cos(x)f''+5f'+f=0, \sin(x)f''=0$$

We have $\sin(x)f''=0$ if and only if $f''=0$ iff $f(x)=Ax+B$, for some $A,B \in \mathbb R$.

Then:

$\cos(x)f''+5f'+f=0$ iff $5A + (Ax+B) = 0 = 0x + 0$ iff $5A+B=0$, $A=0$ iff $A=B=0$

Therefore, the only function is the zero function.