OurBigBook.comSite Source code

Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 13A / a / i / Solution

Open ends impose

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

. By separation of variables,

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] .

Solved by gpt-5.6-sol high.

 Ancestors (12)