How to show that a differential equation has a periodic orbit?

Question

I've been learning differential equations, but I can't solve this exercise:

Show that the following equation has a periodic orbit:

$$u''-(1-u^2)u'+u^5 = 0$$

Can anyone show me how it is done? I've tried to write it as a system

\begin{array}{lcl} x' & = & y \\ y' & = & (1-x^2)y-x^5 \end{array}

but every step i take past this point leads me to a dead end.

Answer 1

Hint.

Considering instead

$$ x^5 x' = x^5 y\\ y y' = y^2(1-x^2)-x^5 y $$

and adding we have

$$ \frac{d}{dt}\left(\frac 16 x^6+\frac 12 y^2\right) = y^2(1-x^2) $$

so we have that

$$ 0 < |x| < 1\Rightarrow \frac{d}{dt}\left(\frac 16 x^6+\frac 12 y^2\right) > 0\\ |x| > 1 \Rightarrow \frac{d}{dt}\left(\frac 16 x^6+\frac 12 y^2\right) < 0 $$