 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ib/paper-1/12d/ii/solution

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

y = e^{- x} v

gives

- v'' + v = λ v

, hence

λ_{n} = n^{2} π^{2} + 1

and

y_{n} = e^{- x} sin (nπ x)

- (e^{2 x} y')' = λ e^{2 x} y

, so

\int_{0}^{1} e^{2 x} y_{n} y_{m} = 0

for

n \neq = m

. Expanding the source gives coefficients

c_{n} = 2 \int_{0}^{1} f sin (nπ x) d x

, hence

a_{n} = \frac{2 [ 1 + ( - 1 ) ^{n} ]}{nπ ( n ^{2} π ^{2} + 1 )} .

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