 Solution

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

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

For

0 \leq t \leq τ

, the delayed argument lies in the prescribed history, so

n^{'} (t) = r

and

n (t) = 1 + r t

. For

τ \leq t \leq 2 τ

n^{'} (t) = r n (t - τ) = r {1 + r (t - τ)} .

Integrating from

τ

and using

n (τ) = 1 + r τ

gives

n (t) = 1 + r t + \frac{r ^{2}}{2} (t - τ)^{2}, τ \leq t \leq 2 τ .

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