Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 6 / Solution

Separate variables near the nonzero initial condition: $y^{-2}dy=dx$. Integration gives $-1/y=x+C$, and $y(1)=1$ fixes $C=-2$. ThereforeThis is the maximal connected interval containing the initial point. Direct differentiation gives $y^{\prime }=1/(2-x{)}^2=y^2$; the pole at $x=2$ prevents continuation as a finite classical solution of a differential equation through that point.