 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-3/13c/a/iii/solution

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

The normalized ansatz intended in the question is

ϕ (s, t) = exp ((s - 1) f (t)),

for which

ϕ (1, t) = 1

. Substitution into the equation from part (ii) and cancellation of

(s - 1) ϕ

f^{'} = λ - β f .

Thus

f (t) = \frac{λ}{β} + (f (0) - \frac{λ}{β}) e^{- βt} .

f (t)

. Equivalently, differentiating at

s = 1

gives

⟨ n ⟩ = \partial_{s} ϕ (1, t) = f (t), σ^{2} = \partial_{s}^{2} ϕ (1, t) + \partial_{s} ϕ (1, t) - \partial_{s} ϕ (1, t)^{2} = f (t) .

Consequently

⟨ n ⟩ ⟶ \frac{λ}{β}, σ^{2} ⟶ \frac{λ}{β} .

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