 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-1/4d/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

For

L (x, y, y^{'}) = y^{'2} - y^{2} - 2 y sin x,

\frac{d}{d x} \frac{\partial L}{\partial y ^{'}} - \frac{\partial L}{\partial y} = 0.

Here this becomes

2 y^{''} + 2 y + 2 sin x = 0, or y^{''} + y = - sin x .

Because the forcing is resonant with the complementary solution, a particular integral is

(x /2) cos x

. Hence

y = A cos x + B sin x + \frac{x}{2} cos x,

and

y^{'} = - A sin x + B cos x + \frac{1}{2} cos x - \frac{x}{2} sin x .

y^{'} (0) = B + \frac{1}{2} = 0, y^{'} (π /2) = - A - \frac{π}{4} = 0.

Thus

y (x) = - \frac{π}{4} cos x - \frac{1}{2} sin x + \frac{x}{2} cos x .

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