 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-2/6c/iv/solution

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

Now

\dot{U} = - ∣\nabla U ∣^{2} + U_{y} cos ω t

, whose second term has either sign. Near the minimum,

ξ = C e^{- 6 t}

and

\overset{η}{˙} + 4 η = cos ω t

, so

η = C_{2} e^{- 4 t} + \frac{4 c o s ω t + ω s i n ω t}{16 + ω ^{2}} .

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