$$\dot{x}=y+x(1-x^2-y^2)(4-x^2-y^2)$$ $$\dot{y}=-x+y(1-x^2-y^2)(4-x^2-y^2)$$
In polar coordinates it's
$$\dot{r}=r(1-r^2)(4-r^2)$$ $$\dot{\theta}=-1$$
I don't understand how the system is converted into the polar coordinates. I know that $r^2=x^2+y^2$.
$r^2 = x^2 + y^2 \therefore r\dot r = x \dot x + y\dot y$.
We have $\dot x = y+x(1-r^2)(4-r^2)$ and $\dot y = -x+y(1-r^2)(4-r^2)$.
Then $$r\dot r = xy+x^2(1-r^2)(4-r^2)-xy + y^2(1-r^2)(4-r^2) = r^2(1-r^2)(4-r^2)$$
Hence $\dot r = r(1-r^2)(4-r^2)$.
Similarly,
$$\theta = \tan^{-1} \frac{y}{x} \therefore \dot \theta =\frac{\frac{\dot y x - x \dot y}{x^2}}{1 + \left(\frac{y}{x}\right)^2} = \cdots = -1$$