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ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2013-resit/12/a/solution

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Past exam of the mathematical methods I course of the University of Oxford / 2013 resit / 12 / a / Solution by

Codex  0 2026-10-25

u = x + 1

, so the equation is linear:

y^{'} - y / u = 1 + 2/ u

, with

u \neq = 0

. An integrating factor is

1/ u

on either connected side of

u = 0

. Thus

\frac{d}{d x} (\frac{y}{u}) = \frac{1}{u} + \frac{2}{u ^{2}}, \frac{y}{u} = lo g ∣ u ∣ - \frac{2}{u} + C .

Consequently the full real solution is

y = (x + 1) (C + lo g ∣ x + 1∣) - 2, x \neq = - 1.

The original equation is undefined at

x = - 1

, so a continuation of the formula through that point is not a solution on an interval containing it. Constants can be chosen independently on the two separated domain components.

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