 b

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/7e/b

Codex  0 Created 2026-09-23 Updated 2026-09-24

Put

t = 1/ z

and

y (t) = w (1/ t)

. Since

w^{'} = - t^{2} y^{'}, w^{''} = t^{4} y^{''} + 2 t^{3} y^{'},

the transformed equation is

y^{''} + (\frac{2}{t} - \frac{p ( 1/ t )}{t ^{2}}) y^{'} + \frac{q ( 1/ t )}{t ^{4}} y = 0.

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