Consider the following ODE: \begin{cases} y''(x)=-y(x)^3 \\ y(0)=1 \\ y'(0)=0 \end{cases} I have to prove that the solution is periodic.
I showed that the quantity $\frac{1}{2}\left (y'(x))^2+\frac{1}{4}(y(x)\right)^4$ is constant and equal to $\frac{1}{4}$, so both $y$ and $y'$ are bounded whenever they are defined, hence $y$ is well defined for every $x\in\mathbb{R}$. I noticed also that $\tilde{y}(x)=y(-x)$ satisfies the same equation of $y$, so $y$ and $\tilde{y}$ coincide. However, I didn't succeed in proving that $y$ is periodic.
Does anyone know how to do it?