 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/34d/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

F_{i} = \frac{1}{2} (p_{i}^{2} + W_{i}^{2} q_{i}^{2} + a_{i} q_{i}), i = 1, \dots, n .

Then

\sum_{i} F_{i} = H

. Each

F_{i}

involves only

(p_{i}, q_{i})

, so

{F_{i}, F_{j}} = 0

for all

i, j

. Their differentials have disjoint coordinate supports and are independent wherever none of them vanishes, which is an open dense regular set.

\overset{q}{˙}_{i} = p_{i}, \overset{p}{˙}_{i} = - W_{i}^{2} q_{i} - \frac{a _{i}}{2} .

After the shift

Q_{i} = q_{i} + \frac{a _{i}}{2 W _{i}^{2}},

they become

n

\ddot{Q}_{i} + W_{i}^{2} Q_{i} = 0.

Thus the $n$ independent commuting integrals establish complete integrability.

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