Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 2 / 8A / b / ii / Solution

The Jacobian isAt $(1,0)$ its eigenvalues are $1,-4$, so this point is a saddle. At $(-1,0)$ they are $(-3\pm i\sqrt{7})/2$, so this point is a stable spiral.

Solved by gpt-5.6-sol high.