Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 14 / c / ii

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- Solution ii

Solution

For

p \neq = 0, \pm 1

, separate variables and use partial fraction decomposition:

\frac{d p}{p ( 1 - p ^{2} )} = - \frac{d x}{2 x}, \frac{1}{p ( 1 - p ^{2} )} = \frac{1}{p} + \frac{p}{1 - p ^{2}} .

Integration yields

lo g ∣ p ∣ - \frac{1}{2} lo g ∣1 - p^{2} ∣ = - \frac{1}{2} lo g ∣ x ∣ + C_{0} .

On any connected solution interval, absorb the fixed signs into a real constant to write

\frac{x p ^{2}}{1 - p ^{2}} = C, p^{2} = \frac{C}{x + C} .

The two real signs of the square root are chosen consistently on a branch where

C / (x + C) > 0

. The constant solutions of the reduced equation are

p = 0, 1, - 1

. The original equation excludes the last two;

p = 0

is retained by

C = 0

for

x \neq = 0

.

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