= Solution
The <nullclines> are $y=\mu x$ and $y=x^2/[\nu(1+x^2)]$. At a nonzero <fixed point>, cancellation of $x$ yields $\mu\nu=x/(1+x^2)$. The latter function attains its maximum $1/2$ at $x=1$. Hence
$$
\boxed{\mu_c=\frac1{2\nu},\qquad
O=(0,0),\quad P_\pm=(x_\pm,\mu x_\pm),\quad
x_\pm=\frac{1\pm\sqrt{1-4\mu^2\nu^2}}{2\mu\nu}.}
$$
For $0<\mu<\mu_c$, $x_-<1<x_+$ and $x_-x_+=1$.
The <Jacobian matrix> is
$$
J(x,y)=\begin{pmatrix}-\mu&1\\2x/(1+x^2)^2&-\nu\end{pmatrix}.
$$
At $O$ the distinct <eigenvalues> are $-\mu,-\nu$, so the origin is a <stable node>. At a nonzero <fixed point> its <trace> is $-(\mu+\nu)<0$ and
$$
\det J=\mu\nu\frac{x^2-1}{1+x^2}.
$$
The discriminant is $(\mu-\nu)^2+8x/(1+x^2)^2>0$. Thus \b[$P_-$ is a saddle and $P_+$ is a <stable node>], never a focus.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ii/paper-1-bistable-flow.png]
{title=Nullclines and bistable phase portrait with saddle manifolds}
Above the straight <nullcline> $\dot x>0$; below it $\dot x<0$. Below the curved <nullcline> $\dot y>0$; above it $\dot y<0$. At the saddle, an <eigenvector> can be chosen as $(1,\mu+\lambda)$: the stable <eigenvalue> gives negative slope, and the unstable <eigenvalue> gives positive slope. The <stable manifold> is the basin boundary. The two branches of the <unstable manifold> lead respectively to the origin and the positive <stable node>, as illustrated for representative parameters.
The nonnegative quadrant is forward invariant. Since $\dot y\leq1-\nu y$, $y$ stays bounded, and then $\dot x=-\mu x+y$ also bounds $x$. Moreover the divergence is the negative constant $-(\mu+\nu)$, excluding periodic or homoclinic loops by the <Bendixson-Dulac criterion>. The <Poincaré-Bendixson theorem> then leaves convergence to a <fixed point>. Initial data on the saddle's <stable manifold> approach the saddle; data in the two open basins approach \b[extinction at $O$ or persistence at $P_+$], respectively.
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