 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/13e/d/solution

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

Take

n = 1

in part (c). The original equation

w_{xx} + w = 0

has the solution

sin x = 1 - z^{2}

. The solution

w_{2}

is normalized by

w_{2} / (1 - z)^{1/2} \to 1

z \to 1

, so

w_{2} (z) = \frac{1}{2} 1 - z^{2} = (1 - z)^{1/2} \frac{1 + z}{2} .

Comparing this with the formula for

w_{2}

and using

u = (1 - z) /2

gives

F (- \frac{1}{2}, \frac{3}{2}; \frac{3}{2}; u) = 1 - u .

This also agrees with the general identity

F (A, C; C; u) = (1 - u)^{- A}

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