 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/31d/b/solution

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

Differentiation under the integral sign gives

U^{'} (x) = - \int_{0}^{\infty} \frac{t e ^{- x t}}{1 + t} d t, U^{''} (x) = \int_{0}^{\infty} \frac{t ^{2} e ^{- x t}}{1 + t} d t .

Hence

x U^{''} + (1 - x) U^{'} - U = \int_{0}^{\infty} e^{- x t} \frac{x t ^{2} - ( 1 - x ) t - 1}{1 + t} d t = \int_{0}^{\infty} e^{- x t} (x t - 1) d t = x \frac{1}{x ^{2}} - \frac{1}{x} = 0.

Thus

x U^{''} (x) + (1 - x) U^{'} (x) - U (x) = 0 .

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