Show that the equation has a periodic orbit I've been learning differential equations and need help with this exercise:
$$
x^{\prime \prime}+\left(5 x^{4}-9 x^{2}\right) x^{\prime}+x^{5}=0
$$
I've tried to write it as a system:
$$
\begin{array}{l}
x^{\prime}=y \\
y^{\prime}=\left(9 x^{2}-5x^4 \right) y-x^{5}
\end{array}
$$
any help would be appreciated
 A: The main idea is that, when you consider a small neighborhood of zero, $x^2$ is the most powerful and $y’$ has the same sign as $y$. So the zero point is unstable. But when $x$ is big enough, higher powers dominate, so trajectories become attracted to zero. The only consensus between those two phenomena is a cycle somewhere in the middle.
Try to find an appropriate Lyapunov function to make it formal. Also, I’m not sure, but such problem should have a known general solution.
A: It is often helpful to combine the first two terms in
$$
x''+c(x)x'+g(x)=0
$$
to form the second variable $y=x'+C(x)$, $C'(x)=c(x)$, in the first order system, which now is
\begin{align}
x'&=y-C(x)\\
y'&=-g(x)
\end{align}
Then you get a phase portrait like

where the blue line is the null-cline $y=C(x)=x^5-3x^3$. This has the typical almost rectangular or trapezoidal structure of a fast-slow system, a Lienard dynamic, ... which is commonly demonstrated on the van der Pol oscillator $\ddot x+\mu(x^2-1)\dot x+x=0$. It is clearly visible that a limit cycle has to exist, however to find a Lyapunov function or to demonstrate that in other ways seems not so easy.
See Trajectory in the Lienard system on how  one can start to show that solutions starting "outwards" have to spiral inwards. Such a spiral then also provides a trapping region.
