 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-4/31c/a/solution

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

1 - λ^{2} / x^{2} = 0

, so the single turning point is

x_{tp} = λ .

Put

p (x) = (λ^{2} / x^{2} - 1)^{1/2}

and

q (x) = (1 - λ^{2} / x^{2})^{1/2}

x = 1

y_{1} (x) \sim A p (x)^{- 1/2} sin (\frac{1}{ϵ} \int_{1}^{x} p (s) d s), 1 \leq x < λ,

while decay at infinity selects

y_{2} (x) \sim Bq (x)^{- 1/2} exp (- \frac{1}{ϵ} \int_{λ}^{x} q (s) d s), x > λ .

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