Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 3 / d

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- Solution d

Solution

Insert the source into the Green-function representation and split the integral at its source discontinuity in derivative:

y (x) = - e^{x} cos x \int_{0}^{x} sin ξ d ξ - e^{x} sin x \int_{x}^{π /2} cos ξ d ξ = - e^{x} [cos x (1 - cos x) + sin x (1 - sin x)] .

Using

sin^{2} x + cos^{2} x = 1

, the answer is

y (x) = e^{x} (1 - cos x - sin x) .

Both endpoint values are zero. As a direct check, conjugating by

e^{x}

sends

y^{''} - 2 y^{'} + 2 y

to

e^{x} (v^{''} + v)

, and

v = 1 - cos x - sin x

gives

v^{''} + v = 1

. Thus the differential equation and both boundary conditions are satisfied.

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