I found in an exercise book the following problem :
Let $y$ be a $1$-periodic solution to $$ y''(x)+a(x)y(x)=0. $$ with $a$ continuous, $a\neq 0$ and $\int_0^1 a(t) \text{d} t = 0$. Prove that every $1$-periodic solution of the previous equation is such that there exists $x\in[0,1]$ such that $y(x)=0$.
I tried to suppose that there exists a solution such that $\forall x\in[0,1]$, $y(x)\neq 0$. I think that in that case, $a$ must be $1$-periodic too. But I don't know where to go next. Any idea?
Suppose that $y(x) \ne 0$ for all $x \in [0,1]$. Then
$-a(x)= \frac{y''(x)}{y(x)}$. Hence
$0=-\int_0^1 a(t) \text{d} t =\int_0^1 \frac{y''(t)}{y(t)} \text{d} t $. Now use integration by parts to obtain
$0=\int_0^1 \frac{y''(t)}{y(t)} \text{d} t=\int_0^1 \frac{y'(t)^2}{y(t)^2} \text{d} t$.
To this end observe that $y(1)=y(0)$ and $y'(1)=y'(0)$.
From $\int_0^1 \frac{y'(t)^2}{y(t)^2} \text{d} t=0$ we get that $y'(x)=0$ for all $x \in [0,1]$, thus $y''(x)=0$ for all $x \in [0,1]$, therefore $a(x)=0$ for all $x \in [0,1]$, a contradiction.