 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-1/2/ii/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

The initial function is odd, so all

a_{n}

, including

a_{0}

, vanish. Set

k_{n} = nπ / L

. For

k_{n} \neq = 1

\int_{0}^{L} sin x sin (k_{n} x) d x = \frac{( - 1 ) ^{n} k _{n} s i n L}{1 - k _{n}^{2}}, \int_{0}^{L} x sin (k_{n} x) d x = - \frac{L ( - 1 ) ^{n}}{k _{n}} .

b_{n} = \frac{2 ( - 1 ) ^{n} sin L}{nπ [ 1 - ( nπ / L ) ^{2} ]} .

Insert them into the sine terms of the preceding solution. If

L = mπ

for a positive integer

m

, the original initial profile is simply

sin x

, with

k_{m} = 1

. The apparent

0/0

in the displayed coefficient has removable limit

b_{m} = 1

, while

b_{n} = 0

for

n \neq = m

. In that case

u (x, t) = e^{- κ t} sin x .

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