 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-2/39c/b/iv/solution

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

f = A + B e^{- C η}

with

C > 0

B C^{2} (A - C) e^{- C η} = 0,

A = C

. The two wall conditions give

A + B = 0, - BC = 1.

Hence

C^{2} = 1

, and positivity of

C

yields

A = 1, B = - 1, C = 1,

f (η) = 1 - e^{- η} .

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