Q: System of differential equations - Equilibria, phase plane, polar coordinates I am currently studying for my exam in ordinary differential equations and I am having a hard time with the following problem:
I have given the system of ODEs:
$$
x^\prime = x+xy-(x+y)\sqrt{x^2+y^2}, \\
y^\prime = y-x^2+(x-y)\sqrt{x^2+y^2}.
$$
I already transformed the system in polar coordinates, which looks like:
$$
r^\prime=r(1-r), \theta^\prime=(1- \text{cos }\theta)r
$$
My first problem: I have to determine the equilibrium points but I don't know how to do it in polar coordinates. How do I get the quilibrium points in polar coordinates and how are they in relation with the system in $x$ and $y$ coordinates.
My second problem: I have to sketch a phase plane and I have no clue how to do it in polar coordinates.
Please help me out.

A: After your conversion to polar coordinates
$$ r^\prime=r(1-r), \theta^\prime=(1- \text{cos }\theta)r $$
Divide to eliminate parameter
$$ \frac{dr}{1-r}= \frac{d \theta}{1-\cos \theta}$$
$$ - \log (1-r) = \cot \theta/2+ \text{constant}$$
$$ (1-r)= c\cdot e^{-cot (\theta/2)}$$
Note that the sign can be either positive or negative for initial/ boundary value point to be either inside limit cycle $r=1,c=0 $ or outside. In the limit  $ \; r\to 1. $
At  $\theta = \pi/4 $ for spirals inside or outside the limit unit circle....
if $  r_i<1 \text{ then } c>0  \text {  and if }\; r_i>1 \text{ then}  \; c<0 $
Points taken are $ (0,0.6), (0,1.25), (0,1.75)$ for limit cycle given below.
Phase portraits for state variables $(r,r')\;, (\theta,\theta')$ can be drawn.

