Given $ \begin{cases} \frac{dr}{dt} = r(1-r),\\ \frac{d\varphi}{dt} = \sin^2\varphi + (1-r)^3, \end{cases}$ I want to draw the phase plane in the $(x,y)$-coordinate system with $x=r\cos(\varphi)$ and $y=r\sin(\varphi)$. I looked at questions that did it the other way round but I don't see how to apply that to this version. I tried substituting $r=\frac{x}{\cos(\varphi)}$ and $r=\frac{y}{\cos(\varphi)}$, but I didn't know where to go from there.
I determined that the stationary points (so both derivatives equal to zero) are $(1,0)$ and $(1,\pi)$ in the $(r,\varphi)$-plane, and therefore $(1,0)$ and $(-1, 0)$ in the $(x,y)$-plane.
I also have a phase plane for the $(r,\varphi)$ plane (in which I drew a few trajectories to check that the stationary points were correct), note that $p=\varphi$ here.
