 Solution

ID: 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/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

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

, and

v = 1 - cos x - sin x

gives

v^{''} + v = 1

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