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ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-iii/1/b/solution

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Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics III / 1 / b / Solution by

Codex  0 2026-10-25

Separation of variables gives

\frac{y d y}{3 y ^{2} + 1} = \frac{x d x}{x ^{2} - 1}, \frac{1}{6} lo g (3 y^{2} + 1) = \frac{1}{2} lo g (x^{2} - 1) + C .

Here

x > 1

makes the argument of the second logarithm positive. Substituting the initial condition yields

3 y^{2} + 1 = 4 (x^{2} - 1)^{3}

. The positive branch is therefore

y (x) = \frac{4 ( x ^{2} - 1 ) ^{3} - 1}{3}, x > 1 + 4^{- 1/3} .

This is the maximal connected interval containing

2

on which this branch solves the original differential equation. At its left endpoint

y = 0

, the original right-hand side has a zero denominator, so one cannot extend the solution through that point merely by continuing the squared equation.

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