= Solution
The <bifurcation diagram> has boundary branches $y=0$ at $(1,0)$ and $y=\mu$ at $(0,\mu)$. The former is stable for $0<\mu<1$ and a saddle for $\mu>1$; the latter is a saddle for $0<\mu<1$ and stable for $\mu>1$. The origin is an unstable equilibrium for every $\mu>0$; its $y=0$ branch coincides in this projection with the $(1,0)$ branch, so it is shown separately in the caption.
The stable interior branch $y_-$ rises from $0$ at $\mu=1$ to $1/2$ at $\mu=5/4$. The unstable interior branch $y_+$ falls from $1$ to $1/2$ over the same interval. These meet at the <saddle-node bifurcation>; at $\mu=1$ they meet the transcritical and pitchfork points respectively. There is bistability between the lower interior node and $(0,\mu)$ for $1<\mu<5/4$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-3-bifurcations.png]
{title=Equilibrium branches and stability for Question 14; the always unstable origin also projects to y equals zero}
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