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

Table of contents
- Solution i

Solution

With

v = x + y

, the chain rule gives

\frac{d v}{d x} = 1 + \frac{d y}{d x} = \frac{2 ( v - 2 )}{3 v - 4} .

For

v \neq = 2

, this separable differential equation gives

x = \int \frac{3 v - 4}{2 ( v - 2 )} d v = \frac{3}{2} v + lo g ∣ v - 2∣ + C .

Restoring

v

and absorbing constants, the nonconstant branches satisfy

x + 3 y + 2 lo g ∣ x + y - 2∣ = C .

The constant branch lost in division is

v = 2

, giving the additional solution

y = 2 - x

. All branches must avoid

v = 4/3

, where the original differential equation is undefined.

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