 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-2/8/a/solution

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

Here

P_{y} = 5 = Q_{x}

, and direct integration constructs a global potential for this exact differential form. Integrate

P

with respect to

x

to obtain

f = 6 x^{2} + 5 x y - 9 x + g (y)

; matching

f_{y} = Q

gives

g^{'} = 3 y - 4

. Thus

f (x, y) = 6 x^{2} + 5 x y - 9 x + \frac{3}{2} y^{2} - 4 y + C .

Both derivatives recover the supplied coefficients.

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