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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/2014/paper-1/13/a/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 / 2014 / Paper 1 / 13 / a / i / Solution by

Codex  0 2026-10-18

An integrating factor is

e^{3 x}

. Multiplication gives

(e^{3 x} y)^{'} = 8 e^{3 x}

, so

y = 8/3 + C e^{- 3 x}

. The initial condition gives

C = 4/3

. Therefore

y (x) = \frac{8}{3} + \frac{4}{3} e^{- 3 x} .

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