 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2/7d/a/solution

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

Substituting

y (z) = \int_{γ} e^{z t} f (t) d t, y^{(3)} (z) = \int_{γ} e^{z t} t^{3} f (t) d t,

and using

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

gives

z y^{(3)} + 2 y = [e^{z t} t^{3} f (t)]_{\partial γ} + \int_{γ} e^{z t} {2 f (t) - (t^{3} f (t))^{'}} d t .

It is therefore enough to require

(t^{3} f)^{'} = 2 f

and choose

γ

so that the integral converges and

e^{z t} t^{3} f (t)

vanishes at its endpoints. Solving the amplitude equation,

\frac{f ^{'}}{f} = \frac{2}{t ^{3}} - \frac{3}{t}, f (t) = C t^{- 3} e^{- 1/ t^{2}} .

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