periodic solution of $x''-\ (1-\ x^2-\ (x')^2)\ x'+x=0$ Assume differential equation $$x''-\ (1-\ x^2-\ (x')^2)\ x'+x=0$$
I want to discusse about non-constant periodic solution of it.
Can someone give a hint that how to start to think. And does it have periodic solution.
My tries: I changed it to system of differential eqution below$$ x_1'=x_2$$ $$x_2'=-x_1+x_2-(x_1^2+x_2^2)x_2$$ I know that if it has perodic solution there exist $t_1$ and $t_2$ such that $x_1(t_1)=0$ and $x_2(t_2)=0$

In terms of $$Z = x^2 + x'^2 - 1$$ the equation becomes
$$Z' = -2x'^2Z$$
which has the solution $Z(t) = Z(0)e^{-2\int_0^t x'^2 dt}$. Now if $x$ is periodic then $Z$ must be periodic, but this is only possible (since $\int_0^t x'^2 dt$ is an increasing function) if $x'\equiv 0$ or $Z(0) = 0$. The only non-constant periodic solutions therefore satisfy
$$Z\equiv 0\iff\frac{x'}{1-x^2} = \pm 1 \iff \arcsin(x) = \pm t + C$$
and we get that all non-constant periodic solutions are on the form $x(t) = \sin(\pm t + C)$ for some constant $C$.
