This is the question in my "homework." I say "homework" because it is not picked up or graded but we are supposed to do it for practice, anyhow here's the question:
Given the system $$x'=-x-y\log^{-1}r, \,\,\,\,\,\,\,y'=-y+x\log^{-1}r,$$ where $r=\sqrt{x^2+y^2}$, show that (0,0) is a simple critical point. Then use the polar form to show that it is a spiral point, whereas it is a proper node for the corresponding linear system.
I understand how it is a proper node for the corresponding linear system, however I'm having trouble understanding how using polar form shows that it is a spiral point. Using polar form the system becomes:
$$r'=-r, \,\,\,\,\,\,\, \theta'=\frac{1}{\log r}.$$
How do I use this to show that (0,0) is a spiral point?
In the polar form, the system $$r'=-r, \quad \theta'=\frac{1}{\log r}$$ can be easily solved explicitly, because the first equation is independent of the second. To wit, $r(t) = Ce^{-t}$. Consequently, $\theta' = \frac{1}{\log C-t}$ There's a singularity at $t=\log C$ (as you can see from the original equations, the ODE does not handle $r=1$ well). For $t>\log C$ the trajectory spirals toward the origin. For $t<\log C$ it spirals toward infinity. I colored the two parts differently on the image below.