Solution
= Solution
The reaction is $F(n)=-n(n-A)(n-B)$. Its homogeneous equilibria are $0,A,B$; $F'(0)<0$ and $F'(B)<0$, while $F'(A)>0$. Thus the stable equilibria are $0$ and $B$.
= Solution
The reaction is $F(n)=-n(n-A)(n-B)$. Its homogeneous equilibria are $0,A,B$; $F'(0)<0$ and $F'(B)<0$, while $F'(A)>0$. Thus the stable equilibria are $0$ and $B$.