 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-4/14c/b/i/solution

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

Substitution gives

(g \tilde{n})^{'} = - (μ + r) \tilde{n}

, hence

\tilde{n} (s) = \frac{C}{g ( s )} exp [- \int_{0}^{s} \frac{μ ( u ) + r}{g ( u )} d u] .

1 = \int_{0}^{\infty} \frac{b ( s )}{g ( s )} exp [- \int_{0}^{s} \frac{μ ( u ) + r}{g ( u )} d u] d s .

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