 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-3/9b/c/i/solution

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

\nabla^{2} ϕ = \frac{1}{r ^{2}} (r^{2} ϕ^{'})^{'} .

Here regularity at zero selects

r^{2} ϕ^{'} (r) = \int_{0}^{r} s^{2} (s - 1) e^{s} d s = e^{r} (r^{3} - 4 r^{2} + 8 r - 8) + 8.

A further integration gives

ϕ (r) = \frac{e ^{r} ( r ^{2} - 5 r + 8 ) - 8}{r} + C,

whose apparent singularity is removable, with limit

3 + C

at zero. Since the nonconstant term has value

4 e - 8

r = 1

ϕ (1) = 3

gives

C = 11 - 4 e

. Therefore

ϕ (r) = \frac{e ^{r} ( r ^{2} - 5 r + 8 ) - 8}{r} + 11 - 4 e .

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