<s>[INST] Prove that $\ddot{x}+(2x^2+\dot{x}^2-1)\dot{x}+x=0$ has a periodic solution. Prove that $\ddot{x}+(2x^2+\dot{x}^2-1)\dot{x}+x=0$ has a periodic solution.
Here's my partial solution: Let $\dot{x}=y$. Then $\ddot{x}=\dot{y}=-(2x^2+y^2-1)y-x=-2x^2y-y^3+y-x$.
So we have $\dot{x}=y$ and $\dot{y}=-2x^2y-y^3+y-x$.
Convert to polar coordinates:
$r\dot{r}=x\dot{x}+y\dot{y}=xy+y(-2x^2y-y^3+y-x)=xy-2x^2y^2-y^4+y^2-xy$.
$r\dot{r}=-2r^4cos^2\theta sin^2\theta-r^4sin^4\theta +r^2sin^2\theta$ which reduces to
$\dot{r}=-2r^3cos^2\theta sin^2\theta-r^3sin^4\theta +rsin^2\theta$.
Now I don't know how to proceed to deduce that we have a periodic solution. Any ideas?
 [/INST] Hint.
$$
\cases{
\dot x_1 = x_2\\
\dot x_2 = -x_1-(2x_1^2+x_2^2-1)x_2
}
$$
or
$$
\cases{
\dot x_1 x_1= x_1x_2\\
\dot x_2 x_2= -x_1x_2-(2x_1^2+x_2^2-1)x_2^2
}
$$
adding the equations
$$
\frac 12(x_1^2+x_2^2)' = -(2x_1^2+x_2^2-1)x_2^2
$$
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