 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/14b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Insert

w (z) = \int_{γ} e^{z t} f (t) d t

into the equation. Since

w^{'} = \int_{γ} t e^{z t} f d t

and

w^{''} = \int_{γ} t^{2} e^{z t} f d t

, use

z e^{z t} = \partial_{t} e^{z t}

z w^{''} + 2 a w^{'} + z w = [e^{z t} (1 + t^{2}) f (t)]_{\partial γ} + \int_{γ} e^{z t} [2 a t f - \frac{d}{d t} ((1 + t^{2}) f)] d t .

The integral vanishes when

(1 + t^{2}) f^{'} = 2 (a - 1) t f,

so, up to a constant and on a consistently chosen branch,

f (t) = (1 + t^{2})^{a - 1} .

The remaining contour condition is

[e^{z t} (1 + t^{2})^{a}]_{\partial γ} = 0 .

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