 Solution

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/a/solution

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

With

M = 2 x y^{2} + 4

and

N = 2 x^{2} y - 6

, the cross derivatives agree:

M_{y} = N_{x} = 4 x y

M

with respect to

x

gives a potential

x^{2} y^{2} + 4 x + g (y)

; matching

N

gives

g^{'} = - 6

x^{2} y^{2} + 4 x - 6 y = C .

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