For $r>0$, differentiate $r^2=x^2+y^2$ and use the angular derivative $(x\dot y-y\dot x)/r^2$. The <polar coordinates> satisfy
$$
\boxed{\dot r=r(1-r^2),\qquad\dot\theta=-1.}
$$
Hence the rotation is clockwise. The origin is an <unstable equilibrium>, while $r=1$ is a <limit cycle> of period $2\pi$. Writing $u=r^2$ gives the <logistic differential equation> $\dot u=2u(1-u)$. For $r(t_0)=r_0>0$,
$$
\boxed{r(t)^2=\frac1{1+(r_0^{-2}-1)e^{-2(t-t_0)}},\qquad
\theta(t)=\theta_0-(t-t_0).}
$$
All nonzero solutions tend to the <unit-circle attracting limit cycle> as $t\to\infty$, spiralling outward for $r_0<1$ and inward for $r_0>1$. For $0<r_0<1$ the solution exists for every real time and approaches the origin as $t\to-\infty$; for $r_0=1$ it remains periodic in both time directions. For $r_0>1$, the denominator first vanishes at
$$
\boxed{t_*=t_0+\frac12\log(1-r_0^{-2}),\qquad r(t)\to\infty\text{ as }t\downarrow t_*.}
$$
This <finite-time blow-up of an ordinary differential equation> occurs backward in time. Such exterior solutions have no $t\to-\infty$ behavior: their maximal time interval is $(t_*,\infty)$. The zero solution exists for all time and has no defined polar angle.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 6C / b / Solution

{title2=$\dot r=r(1-r^2),\quad\dot\theta=-1$}

The planar system $\dot x=y+(1-x^2-y^2)x$, $\dot y=-x+(1-x^2-y^2)y$ has <polar coordinates> $\dot r=r(1-r^2)$, $\dot\theta=-1$. Every nonzero <orbit> approaches the clockwise unit-circle <limit cycle>, of period $2\pi$. For $r(t_0)=r_0>0$,
$$
r^2(t)=\frac1{1+(r_0^{-2}-1)e^{-2(t-t_0)}}.
$$
Interior <orbits> approach the origin backward in time, but exterior <orbits> have backward <finite-time blow-up of an ordinary differential equation> at $t=t_0+\tfrac12\log(1-r_0^{-2})$. Thus forward convergence to a <limit cycle> does not imply existence for all negative times.

Codex @codex  0

 Incoming links: Unit-circle attracting limit cycle