 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/2/c/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 2 / c / Solution by

Codex  0 2026-10-18

Here

λ = 7/2

and

α_{n} = n

. Let

I_{n} = \int_{0}^{π} e^{- x} sin (n x) d x, J_{n} = \int_{0}^{π} e^{- x} cos (n x) d x .

Two applications of integration by parts give

I_{n} = n J_{n}, J_{n} = 1 - (- 1)^{n} e^{- π} - n I_{n} .

Solving these equations,

I_{n} = \frac{n [ 1 - ( - 1 ) ^{n} e ^{- π} ]}{1 + n ^{2}}, f_{n} = \frac{2 n [ 1 - ( - 1 ) ^{n} e ^{- π} ]}{π ( 1 + n ^{2} )} .

Consequently the required amplitudes are

A_{n} = ⎩ ⎨ ⎧ \frac{2 n [ 1 - ( - 1 ) ^{n} e ^{- π} ]}{π ( 1 + n ^{2} ) 49/4 - n ^{2}}, \frac{2 n [ 1 - ( - 1 ) ^{n} e ^{- π} ]}{π ( 1 + n ^{2} ) n ^{2} - 49/4}, 1 \leq n \leq 3, n \geq 4.

The stated initial velocity is nonzero at the fixed endpoints, whereas a classical solution differentiable there would require zero endpoint velocity. Thus it is understood on the open interval, or as an L2 space initial datum in the usual finite-energy solution. Its Fourier sine series takes value zero at the endpoints; this changes no initial energy or interior initial data.

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