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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/2006/paper-2/7/b/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 / 2006 / Paper 2 / 7 / b / Solution by

Codex  0 Created 2026-10-18 Updated 2026-10-20

Here

M_{y} = 2 y

whereas

N_{x} = 0

, so the equation is not exact. The identity

M_{y} - N_{x} = N

gives the integrating factor for a differential one-form

μ (x) = e^{x}

. Multiplying makes the left-hand side the exact differential of

e^{x} (y^{2} - x + 1)

. Thus

e^{x} (y^{2} - x + 1) = C, or y^{2} = x - 1 + C e^{- x} .

(1)

For real graph solutions, the right-hand side must be nonnegative; a smooth choice of square-root branch is made on the interval concerned. Finding the factor did not require dividing by

y

at zero.

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