 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 isThe 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.

The local bifurcation curves areAt $(0,0)$ the linearization has a double zero eigenvalue and is nilpotent, giving a Bogdanov–Takens bifurcation. For $\mu \ne 0$ 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.