 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-2/5a/b/solution

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

For

x > 0

x

gives

y^{'} - (2 x + x^{- 1}) y = x .

e^{- x^{2}} / x

, so

y = x e^{x^{2}} (C + \int_{0}^{x} e^{- t^{2}} d t) .

Every choice except

C = - \int_{0}^{\infty} e^{- t^{2}} d t = - \frac{π}{2}

grows like

x e^{x^{2}}

y_{b} (x) = - x e^{x^{2}} \int_{x}^{\infty} e^{- t^{2}} d t .

Integration by parts, or the Gaussian-tail asymptotic, gives

\int_{x}^{\infty} e^{- t^{2}} d t \sim \frac{e ^{- x^{2}}}{2 x}, y_{b} (x) \to - \frac{1}{2} .

It starts from

0

x = 0

, remains negative, and decreases monotonically toward the horizontal asymptote

- 1/2

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