Finding a good Lyapunov function for $\{ x'= y , y'= -4 x + 5x^3 - x^5 \}$

For the system:

$$\begin{cases}x'= y \\y'= -4 x + 5x^3 - x^5 \end{cases}$$

I am trying to determine the stability of $$(x,y)=(0,0)$$ by means of a Lyapunov function. I am trying to find a good one, the regular $$V(x,y)=ax^2 + by^2$$ does not help me as I get odd-powered terms and products of $$x$$ and $$y$$ that do not cancel. Specifically: $$\dot{V}(x,y) = 2axy + b xy(-1+5x^2-x^4)$$
Does someone have a better suggestion, what is the general approach in finding such a function for a given problem? I want to somehow use the fact that these odd powers of $$x$$ and $$y$$ appear in this system of equations, I haven't figured out how to do this in an effective manner.

Hint.

The dynamical system has an integral which is

$$\frac 12 y^2+\frac{x^6}{6}-\frac{5 x^4}{4}+2 x^2=C$$

Studying the level curves we have the following graphics and for $$0 < C < 0.915$$ we have closed level curves around the origin characterizing a center.

• If I use this function as a Lyapunov function I get that $\dot{V}=0$ which means we have a stable solution. This corresponds with the other explanation by Hans and your level curves. Funny how this is a conservative system. – Wesley Strik Jun 20 at 19:42

This is a conservative system, since it's equivalent to $$x'' = -4x + 5x^3 - x^5 = -V'(x) ,$$ with the potential $$V(x) = 2x^2 - \tfrac54 x^4 + \tfrac16 x^6 .$$ So $$H(x,y)=\tfrac12 y^2 + V(x)$$ is a constant of motion, and the trajectories follow the level curves of $$H$$, which near the origin look like ellipses $$2 x^2 + \tfrac12 y^2 = C$$. This means that the origin is neutrally stable (i.e., stable but not asymptotically stable).

Compare the two plots (on Wolfram Alpha):