 Solution

The second equation gives $y=x(1-x)$. The first then says either $y=0$, giving the persistent equilibrium points $(0,0)$ and $(1,0)$, orgiving $x_{\pm }=((\beta +1)\pm \sqrt{(\beta +1{)}^2+4\mu })/2$ and $y_{\pm }=x_{\pm }(1-x_{\pm })$. These two points exist for $\mu \ge -(\beta +1{)}^2/4$.

At the origin the Jacobian matrix has trace $-1$ and determinant $-\mu$, so it is stable for $\mu <0$ and a saddle point for $\mu >0$. At $(1,0)$ the trace is $-1$ and determinant $\mu +\beta$, so it is stable for $\mu >-\beta$ and a saddle for $\mu <-\beta$. Stationary bifurcations occur atwith the first two coinciding when $\beta =1$.