 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/33a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

J (x, y) = (μ - y^{2} - 6 x^{2} - 2 x y - 2 x y 1 - x^{2} - 3 y^{2}) .

(0, 1)

J (0, 1) = diag (μ - 1, - 2),

so one eigenvalue crosses zero at

μ = 1

while the other remains negative. Thus a bifurcation occurs there.

On the

x

-axis the nonzero fixed point is

(x, y) = (\frac{μ}{2}, 0) .

- 2 μ, 1 - \frac{μ}{2} .

The second crosses zero at

μ_{c} = 2,

which is the second bifurcation value.

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