 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-3/10a/b/i/solution

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

\nabla^{2} h = h^{''} + 2 h^{'} / r

. Since

\nabla^{2} e^{- r} = e^{- r} - 2 e^{- r} / r

and

\nabla^{2} r^{- 4} = 12 r^{- 6}

, decay and the boundary value give

u = e^{- r} + r^{- 4} - \frac{e ^{- 1}}{r} .

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