 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/16/a/i/solution

P d x + Q d y = 0

, with

P = 3 x + y

and

Q = 2 x

. Then

\frac{P _{y} - Q _{x}}{Q} = \frac{1 - 2}{2 x} = - \frac{1}{2 x},

μ = ∣ x ∣^{- 1/2}

on either half-line. For example, on

x > 0

the resulting exact form is

(3 x + \frac{y}{x}) d x + 2 x d y = d [2 x (x + y)] .

Hence the solution, written to cover either sign of

x

, is

y = - x + \frac{C}{∣ x ∣}, x \neq = 0 .

The original equation remains meaningful at

x = 0

, where it requires

y = 0

. Only

C = 0

gives a finite differentiable extension across zero, namely

y = - x

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