 Solution

At an interior equilibrium point, $x^2=1-y^2=\mu -y$, soFor $1<\mu <5/4$ both values satisfy $0<y<1$, giving two physical equilibria. At such an equilibrium,The branch $y_{-}<1/2$ is asymptotically stable by the trace-determinant stability criterion; $y_{+}>1/2$ is a saddle equilibrium. In fact $J$ is symmetric, so the stable branch has two negative real eigenvalues.

At $\mu =5/4$ the equilibria merge at $(x,y)=(\sqrt{3}/2,1/2)$. Writing $y=1/2+s$, the equilibrium condition is $s^2+\mu -5/4=0$, the characteristic branch geometry of a saddle-node bifurcation. The stable node and saddle annihilate as $\mu$ increases through $5/4$.