 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-2/5a/a/solution

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

Let

I (x) = exp (\int^{x} p (s) d s) .

(I y)^{'} = I f

gives

y (x) = I (x)^{- 1} (C + \int^{x} I (t) f (t) d t) .

y_{1}, y_{2}

are distinct solutions, their difference is a nonzero homogeneous solution, so every solution is

y = y_{1} + C (y_{2} - y_{1}) .

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