 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3/13a/a/i/solution

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

Open ends impose

C (0, t) = C (L, t) = 0

C (x, t) = \sum_{n = 1}^{\infty} b_{n} sin \frac{nπ x}{L} exp [(μ - \frac{D n ^{2} π ^{2}}{L ^{2}}) t] .

Expanding the initial constant gives

b_{n} = 2 C_{0} (1 - (- 1)^{n}) / (nπ)

, hence

C (x, t) = \frac{4 C _{0}}{π} \sum_{n \geq 1 n m o dd} \frac{1}{n} sin \frac{nπ x}{L} exp [(μ - \frac{D n ^{2} π ^{2}}{L ^{2}}) t] .

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