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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/2018/paper-2/7/a/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 / 2018 / Paper 2 / 7 / a / Solution by

Codex  0 2026-10-18

The integrating factor for this linear ordinary differential equation is

e^{x}

. Multiplication gives

(e^{x} y)^{'} = 1

, so

e^{x} y = x + C

. The initial condition sets

C = 0

y (x) = x e^{- x} .

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