= Solution
The <equilibrium points> satisfy $y=0$, $x^2=\lambda$. For $\lambda<0$ there are none. For $\lambda>0$ set $a=\sqrt\lambda$. The <Jacobian matrix> is
$$
L(x_*)=\begin{pmatrix}0&1\\2x_*&\mu+x_*\end{pmatrix},\qquad\det L=-2x_*,\quad\operatorname{tr}L=\mu+x_*.
$$
The positive <equilibrium point> $(a,0)$ is always a <saddle equilibrium>. At $(-a,0)$ the <determinant> is $2a$, so it is stable for $\mu<a$ and unstable for $\mu>a$. The discriminant $(\mu-a)^2-8a$ distinguishes a node from a focus; its zero does not destroy hyperbolicity and is not another local bifurcation curve.
\b[The local bifurcation curves are]
$$
\boxed{\lambda=0\text{ (saddle-node, }\mu\ne0\text{)},\qquad\mu=\sqrt\lambda\text{ (Hopf, }\lambda>0\text{)}.}
$$
At $(0,0)$ the <linearization> has a double zero <eigenvalue> and is nilpotent, giving a <Bogdanov–Takens bifurcation>. For $\mu\ne0$ on $\lambda=0$, solving the local <center manifold> gives $\dot x=(\lambda-x^2)/\mu+\cdots$, displaying the <saddle-node bifurcation>. For $\mu<0$ the nonsaddle branch born at positive $\lambda$ is attracting; for $\mu>0$ it is repelling. On the Hopf curve the angular <frequency> is $\sqrt{2a}$. The permitted <subcritical Hopf bifurcation> creates a repelling <periodic orbit> on the stable-focus side $\mu<a$.
A local analysis alone does not say how far that <periodic orbit> extends. The global calculation below supplies the additional division of the stable-focus region. The complete nearby <phase portraits> and the leading global curve are shown together here. The $lambda<0$ flow has no closed <periodic orbit>, since a closed orbit would have index one and hence enclose an <equilibrium point>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/iii/paper-51-bogdanov-takens-portraits.png]
{title=Local and saddle-loop bifurcation curves with four nearby phase portraits}
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