Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 32B / b / Solution

The Jacobian matrix isAt $(0,0)$ its eigenvalues are $-k$ and $-1$. It is therefore a stable node for $k>0$, a saddle for $k<0$, and nonhyperbolic at $k=0$, consistent with the proposed transcritical bifurcation.

At $(0,1)$ the matrix iswith eigenvalues $1-k$ and $1$. It is an unstable node for $k<1$, a saddle for $k>1$, and nonhyperbolic at $k=1$, again consistent with a transcritical bifurcation.

Solved by gpt-5.6-sol high.