Finding all twice differentiable $2 \pi$ periodic functions satisfying $e^{ix}f''(x)+5f'(x)+f(x)=0$

Question

I need to find all the functions $f : \Bbb R \to \Bbb R$ , twice continuously differentiable with period $2\pi$ such that: $$e^{ix}f''(x)+5f'(x)+f(x)=0$$

Using the formula $\hat f'(n)=in \hat f(n)$, I got: $$\hat f(n) +5in \hat f(n)-(n-1)^2 \hat f(n-1)=0$$

I concluded that $\hat f(n)=0$ for every $n>0$. But what about the negative coefficients?

Answer 1

From $e^{ix}f''(x)+5f'(x)+f(x)=0$ we get

$\cos x f''(x)+5f'(x)+f(x)+i \sin x f''(x)=0$. Since $f$ is real-valued, we derive

$\sin x f''(x)=0$.

For $x \in (0,2 \pi) \setminus \{\pi\}$ we then have $f''(x)=0$, hence $f$ has the form $f(x)=ax+b$ .

By continuity: $f(x)=ax+b$ for all $x$.

Since $f$ is $2 \pi$ - periodic: $a=0$.

From $e^{ix}f''(x)+5f'(x)+f(x)=0$ we then get: $b=0$