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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/2010/paper-1/19/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 / 2010 / Paper 1 / 19 / a / Solution by

Codex  0 2026-10-18

P d x + Q d y = d F

with continuous second partial derivatives of

F

, then

P = F_{x}

Q = F_{y}

. The symmetry of second derivatives requires

P_{y} = Q_{x} .

(1)

This is a necessary condition for an exact differential. With continuously differentiable coefficients it is also locally sufficient, and is globally sufficient on a simply connected open domain; a domain with holes can require additional global information.

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