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

Separation of variables givesHere $x>1$ makes the argument of the second logarithm positive. Substituting the initial condition yields $3y^2+1=4(x^2-1{)}^3$. The positive branch is thereforeThis is the maximal connected interval containing $\sqrt{2}$ 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.