 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/1a/b/solution

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

Multiplying

y^{'} + P (x) y = Q (x) y^{n}

y^{- n}

and setting

z = y^{1 - n}

gives

\frac{1}{1 - n} z^{'} + P z = Q .

z^{'} + (1 - n) P z = (1 - n) Q .

For the stated equation,

\overset{x}{˙} + 6 t^{2} x = 2 t e^{- t^{3}} x^{1/2} .

z = x

. Then

\overset{z}{˙} + 3 t^{2} z = t e^{- t^{3}} .

e^{t^{3}}

, so

\frac{d}{d t} (e^{t^{3}} z) = t .

Hence

e^{t^{3}} x = \frac{t ^{2}}{2} + C .

The condition

x (1) = 4

gives

C = 2 e - \frac{1}{2}

. Consequently

x (t) = e^{- 2 t^{3}} (\frac{t ^{2}}{2} + 2 e - \frac{1}{2})^{2} .

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