 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-4/31a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

y^{''} + p (x) y^{'} + q (x) y = 0

p (x) = O (x^{- 1})

and

q (x) = O (x^{- 2})

. Here

p = - 1

and

q = - 2/ x - 2/ x^{2}

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