 Solution

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

Codex  0 Created 2026-09-24 Updated 2026-10-05

The fixed point branches and their stability are as follows.

$(0, 0)$ has eigenvalues $μ$ and $1$ , so it is unstable for every $μ > 0$ .
$(0, 1)$ has $x = 0$ and is stable for $0 < μ < 1$ , then a saddle for $μ > 1$ .
$(μ /2, 0)$ exists for all $μ > 0$ ; it is a saddle for $0 < μ < 2$ and stable for $μ > 2$ .
$(μ - 1, 2 - μ)$ exists for $1 < μ < 2$ . Its Jacobian matrix is

J = - 2 (2 x^{2} x y x y y^{2}),

which is a negative-definite matrix, so this branch is stable.

Thus the

x

-versus-

μ

x = 0

branch up to

μ = 1

, a stable branch

x = μ - 1

joining the bifurcation points

(1, 0)

and

(2, 1)

, and the branch

x = μ /2

changing from unstable to stable at

μ = 2

. The origin branch also lies at

x = 0

but remains unstable throughout.

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