= Solution
The second equation gives $y=x(1-x)$. The first then says either $y=0$, giving the persistent equilibrium points $(0,0)$ and $(1,0)$, or
$$
x^2-(β+1)x-μ=0,
$$
giving $x_±=((β+1)±\sqrt{(β+1)^2+4μ})/2$ and $y_±=x_±(1-x_±)$. These two points exist for $μ\geq-(β+1)^2/4$.
At the origin the <Jacobian matrix> has trace $-1$ and determinant $-μ$, so it is stable for $μ<0$ and a <saddle point> for $μ>0$. At $(1,0)$ the trace is $-1$ and determinant $μ+β$, so it is stable for $μ>-β$ and a saddle for $μ<-β$. Stationary bifurcations occur at
$$
μ=-{(β+1)^2\over4},\qquad μ=-β,\qquad μ=0,
$$
with the first two coinciding when $β=1$.
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