Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 13 / b / i

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- Solution i

Solution

Let

P = μ (x) (x y - 16 x - x^{3})

and

Q = μ (x) (16 + x^{2})

. An exact differential form must satisfy

P_{y} = Q_{x}

. These partial derivatives give

xμ = (16 + x^{2}) μ^{'} + 2 xμ, \frac{μ ^{'}}{μ} = - \frac{x}{16 + x ^{2}} .

Taking antiderivatives produces

ln ∣ μ ∣ = - \frac{1}{2} ln (16 + x^{2}) + C

. A convenient nonzero integrating factor for a differential one-form is therefore

μ (x) = \frac{1}{16 + x ^{2}} .

An explicit potential, which also proves global exactness, is

F (x, y) = y 16 + x^{2} - \frac{1}{3} (16 + x^{2})^{3/2} .

Indeed

F_{x} = P

and

F_{y} = Q

for the chosen integrating factor.

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