 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/6a/c/solution

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

ω = π / (2 τ)

. The condition

r τ = - π /2

says

r = - ω

. Then

n (t) = C cos (ω t)

satisfies

n^{'} (t) = - C ω sin (ω t), r n (t - τ) = - C ω cos (ω t - π /2) = - C ω sin (ω t) .

It is periodic with period

2 π / ω = 4 τ

. Every nonzero such solution is negative during half of each period, so it cannot represent a population size.

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