 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-2/5c/b/solution

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

x^{2}

W = C / x

y_{1} = x^{- 1/2} e^{- x}

allows the convenient choices

y_{2} = x^{- 1/2} e^{x}, W = \frac{2}{x} .

The normalized forcing is

f = x^{3/2}

, so

u^{'} = - \frac{1}{2} x^{2} e^{x}, v^{'} = \frac{1}{2} x^{2} e^{- x} .

Taking convenient antiderivatives and simplifying gives the particular solution

- x^{3/2} - 2 x^{- 1/2}

. Hence

y = C_{1} x^{- 1/2} e^{- x} + C_{2} x^{- 1/2} e^{x} - x^{3/2} - 2 x^{- 1/2} .

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