 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-4/31e/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

x^{2}

, the equation becomes

y^{''} + \frac{1}{x} y^{'} - \frac{1}{x ^{4}} y = 0.

x (1/ x)

and

x^{2} (- 1/ x^{4})

to be analytic at zero. The second expression is

- x^{- 2}

, so the criterion fails:

x = 0 is an irregular singular point .

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