 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-3/3b/b/ii/solution

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

On the annulus the logarithmic term is allowed, so

u (r) = r^{4} + C lo g r + D .

The condition

u (1) = 1

gives

D = 0

, while

u (2) = 1

gives

16 + C lo g 2 = 1.

Therefore

u (r) = r^{4} - \frac{15 lo g r}{lo g 2} .

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