$$x = r\cos(\theta);$$ $$y = r \sin(\theta);$$ $$r^2 = x^2 + y^2$$
We differentiate to get $$2rr' = 2xx' + 2yy'$$ -> which simplifies to -> $$rr' = xx' +yy'$$
We plug in $r'$ which is given, then I have $r^2(|r-2|)(r-3) = r \cos(\theta)x' + r\sin(\theta)y'$
But I have not been given an equation with $x$ or $y$ so how can I solve this problem?
Thank You
You do not have to convert to rectangular coordinates to solve your problem.
Note that you have three values for $r$ which make $r'=0$
These values are $r=0, r=2, r=3$
$r=0$ is the origin which is an equilibrium point. This equilibrium point is stable as you can figure it out by looking at $r'$ near $r=0$
$r=2$ is a limit cycle. This limit cycle is unstable as you can check the sign of $r'$ for values of $r<2$ and close to $r=2$
$r=3$ is a limit cycle. This limit cycle is also unstable as you can check the values of $r'$ near $r=3$