 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/31d/c/solution

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

x

, the equation is

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

y = e^{- \frac{1}{2} \int (1/ x - 1) d x} v = e^{x /2} x^{- 1/2} v .

With

P = \frac{1}{x} - 1, Q = - \frac{1}{x},

the normal-form coefficient is

Q - \frac{1}{2} P^{'} - \frac{1}{4} P^{2} = - \frac{1}{4} - \frac{1}{2 x} + \frac{1}{4 x ^{2}} .

Consequently

v^{''} - \frac{1}{4} (1 + \frac{2}{x} - \frac{1}{x ^{2}}) v = 0,

so the constants in the PDF's expression

1 + A x^{- 1} + B x^{- 2}

are

A = 2, B = - 1 .

To classify infinity, put

z = 1/ x

. Since

\frac{d ^{2} v}{d x ^{2}} = z^{4} v_{zz} + 2 z^{3} v_{z},

v

in the transformed equation has a pole of order four at

z = 0

. This exceeds the order two allowed at a regular singular point. Hence

x = + \infty

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