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

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- Solution a

Solution

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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