= Solution
The <fixed point> branches and their stability are as follows.
* $(0,0)$ has <eigenvalues> $\mu$ and $1$, so it is unstable for every $\mu>0$.
* $(0,1)$ has $x=0$ and is stable for $0<\mu<1$, then a saddle for $\mu>1$.
* $(\sqrt{\mu/2},0)$ exists for all $\mu>0$; it is a saddle for $0<\mu<2$ and stable for $\mu>2$.
* $(\sqrt{\mu-1},\sqrt{2-\mu})$ exists for $1<\mu<2$. Its <Jacobian matrix> is
$$
J=-2\begin{pmatrix}2x^2&xy\\xy&y^2\end{pmatrix},
$$
which is a <negative-definite matrix>, so this branch is stable.
Thus the $x$-versus-$\mu$ <bifurcation diagram> has the stable $x=0$ branch up to $\mu=1$, a stable branch $x=\sqrt{\mu-1}$ joining the bifurcation points $(1,0)$ and $(2,1)$, and the branch $x=\sqrt{\mu/2}$ changing from unstable to stable at $\mu=2$. The origin branch also lies at $x=0$ but remains unstable throughout.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2-bifurcation-diagram.png]
{title=Fixed-point bifurcation diagram for the 2025 Paper II 2 system}
{description=Stable branches are solid blue and unstable or saddle branches are dashed red. The interior branch emerges at mu equals 1 and joins the axial branch at mu equals 2.}
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