Writes in polar coordinates $x''-(1-x^2-(x')^2)x'+4x=0$. I need to determine a periodic solution for :$x''-(1-x^2-(x')^2)x'+4x=0$.
We have the equivalent system:
$$\begin{cases} 
       x'=y \\
       y'=(1-x^2-y^2)y-4x.
   \end{cases}
$$

*

*We determined the stationary points for the equivalent system: $(0,0)$.


*I saw that this solution is unstable.


*But now I have to turn this system into polar coordinates. Unfortunately, I tried in all possibilities and I failed to bring the system to a beautiful shape depending on the polar coordinates.
I present what I tried:
Polar coordinates:
$\begin{cases} 
       x(r,\theta)=r\cos(\theta) \\
       y(r,\theta)=r\sin(\theta)
   \end{cases}
$
We have that $x^2+y^2=r^2$ and $\tan(\theta)=\frac{y}{x}$. So $r'=\frac{xx'+yy'}{r}$ and $\theta'=\frac{xy'-x'y}{r^2}.$
So $$r'=\frac{xy+y[(1-x^2-y^2)y-4x]}{r}=\frac{xy+y^2-x^2y^2-y^4-4xy}{r}.$$
I tried to calculate, but I have no idea how to bring the system into polar coordinates to continue my work.
Thanks!

Note $$\theta'=\frac{(1-x^2-y^2)xy-4x^2-y^2}{r^2}=((1-r^2)\sin\theta-3\cos\theta)\cos\theta-1$$ and $$r'=\frac{xy+(1-x^2-y^2)y^2-4xy}r=-3r\sin\theta\cos\theta+(1-r^2)r\sin^2\theta$$ so $$(\log r)'=((1-r^2)\sin\theta-3\cos\theta)\sin\theta.$$ Thus $$\frac{(\log r)'}{\theta'+1}=\tan\theta.$$
