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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/13/b/i/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 13 / b / i / Solution by

Codex  0 2026-10-18

For

x \neq = 0

, multiply by

x

to obtain the first-order linear ordinary differential equation

y^{'} + x y = 5 x e^{- x^{2}}

. Its integrating factor is

μ = e^{x^{2} /2}

. The product rule then gives

(y e^{x^{2} /2})^{'} = 5 x e^{- x^{2} /2} .

Integration yields

y e^{x^{2} /2} = - 5 e^{- x^{2} /2} + C

, so

y (x) = C e^{- x^{2} /2} - 5 e^{- x^{2}}, x \neq = 0.

The constant of integration is arbitrary on each connected interval of the original equation's domain. The formula extends smoothly through zero, but the printed equation itself contains

1/ x

and is not defined there.

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