= Solution
For $r>0$, the <polar coordinates> satisfy
$$
\dot r=\frac{x\dot x+y\dot y}{r}=r(1-r^2),
\qquad \dot\theta=\frac{x\dot y-y\dot x}{r^2}=1.
$$
A positive constant radius must therefore be one. The nonconstant <periodic solutions> are
$$
\boxed{x(t)=\cos(t+\theta_0),\qquad y(t)=\sin(t+\theta_0),}
$$
with least <period> $2\pi$ and counterclockwise orientation. The origin is also a stationary solution, hence periodic in the degenerate sense of a constant function; it is a point rather than a nontrivial closed <orbit>. There are no other constant-radius solutions.
For $r_0>0$, put $v=r^2$. Then $\dot v=2v(1-v)$, a <logistic differential equation>. Solving this <separable differential equation> gives
$$
\boxed{r(t)=\left[1+(r_0^{-2}-1)e^{-2t}\right]^{-1/2},
\qquad \theta(t)=\theta_0+t,\qquad t\geq0.}
$$
The denominator is positive for every forward time. When $0<r_0<1$, the radius increases monotonically to one; when $r_0>1$, it decreases monotonically to one; and when $r_0=1$ it stays there. Thus every nonzero <orbit> approaches the unit circle while continuing to rotate. More precisely, its distance from the phase-matched circular solution tends to zero; its Cartesian coordinates do not converge to a single point. For $r_0=0$, the solution stays at the origin.
The origin is the only <equilibrium point>. Its <Jacobian matrix> is
$$
J(0)=\begin{pmatrix}1&-1\\1&1\end{pmatrix},
$$
whose <eigenvalues> are $1\pm i$. It is an unstable <focus>. The unit circle is an attracting <limit cycle>, rather than a curve of equilibria: on it the vector field remains nonzero. Linearizing the radial equation at $r=1$ gives $\dot{\delta r}=-2\delta r+O(\delta r^2)$, confirming radial attraction. All other radii are strictly monotone, so there are no additional nontrivial <periodic orbits>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ia/paper-2-phase-portrait.png]
{title=Counterclockwise spirals approach an attracting unit circle around an unstable equilibrium}
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