Show that a cubic second order ODE only has periodic solutions 
Consider the following ODE $$x''+x+x^3=0$$ Show that it only has periodic solutions.


I tried to do the following, we consider the equivalent system
$$\begin{cases} x' = y\\ y' = -x(1+x^2) \end{cases}$$
Let $(a(t), b(t))$ be a solution of the last ODE and $I \subseteq \mathbb{R}$ the maximum interval in which it is defined, we can assume that $0\in I$. Then consider the IPV given by the ODE and the initial condition $x(0)=a(0); y(0)=b(0)$ since $(y,-x(1+x^2))$ is locally Lipschitz there is a unique solution $(x(t),y(t))$ therefore  $x(t)=a(t);y(t)=b(t),\quad \forall t\in I$.
If $x(0)=y(0)=0$ the solution is trivial so we can assume that $x_0:=x(0)\neq 0 \neq y(0)=:y_0$.
The function $F(x,y)=2(x^2+y^2)+x^4$ is a constant of motion, thus $(x(t),y(t))\in H:=\{(x,y):F(x,y)=F(x_0,y_0)\}$ $\forall t\in I$. It is easy to see that this is a compact set and a Jordan curve, and contains no fixed points since $F(x_0,y_0)>0$. We also that $\gamma^+(x_0,y_0)$ is bounded therefore the limit set is non-empty, compact and connected. Since $H$ is compact we also know that the limit set is contained in $H$.
I'm stuck at this point, I tried to use Poincaré-Bendixson's theorem from which we deduce that the limit set is periodic but I don't know how to continue from there to conclude that the orbit is periodic.
 A: You found that along a solution the expression
$$
\dot x^2+x^2(1+\tfrac12x^2)=R^2
$$
is constant. The left side can be seen a sum of two squares, so that the whole is a circle equation for radius $R$. Parametrize with the angle as $$\dot x=R\cos(u), ~~ x\sqrt{1+\frac12 x^2}=f(x)=R\sin(u).$$ Then taking the derivative of the second expression and comparing with the first gives
$$
f'(x(t))\dot x(t)=R\cos(u(t))\dot u(t)\\
\implies f'(x(t))=\dot u(t)
$$
for any motion where $\dot x$ is not constant zero.
Now $f(x)$ is a strictly increasing function, $f'(x)\ge 1$ is always positive. This means that $u(t)$ is also strictly increasing, and the phase curve $(x(t),\dot x(t))$ moves along the image of a circular motion, thus is periodic.

$$
f'(x)=\sqrt{1+\tfrac12 x^2}+\frac{x^2}{2\sqrt{1+\tfrac12 x^2}}=\frac{1+x^2}{\sqrt{1+\tfrac12 x^2}}\ge 1
$$
and $y=f(x)$ can be solved observing $\operatorname{sgn}(y)=\operatorname{sgn}(x)$ as follows
$$
y^2=x^2+\tfrac12x^4\\
1+2y^2=(1+x^2)^2\\
x^2=\sqrt{1+2y^2}-1=\frac{2y^2}{1+\sqrt{1+2y^2}}\\
\implies
x=\frac{\sqrt2 y}{\sqrt{1+\sqrt{1+2y^2}}}
$$
