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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/2015/paper-2/7/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 / 2015 / Paper 2 / 7 / Solution by

Codex  0 2026-10-18

On

x \geq 1

, an integrating factor is

1/ x

. Multiplication gives

(y / x)^{'} = 1

, so the general solution is

y = x^{2} + C x

. The initial value

y (1) = 1

C = 0

. Therefore

y = x^{2}, x \geq 1 .

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