 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-3/31b/d/solution

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

For

b = 1

A = a + 1/4

and

X = x - 1/2

. The centre manifold has

y = A - X^{2} + \dots

, so

\dot{X} = A - X^{2} + \dots

. This is a saddle-node at

a = - 1/4

, with two branches for

a > - 1/4

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