Solution
= Solution
Separate variables near the nonzero <initial condition>: $y^{-2}dy=dx$. <Integration> gives $-1/y=x+C$, and $y(1)=1$ fixes $C=-2$. Therefore
$$
\boxed{y(x)=\frac1{2-x},\qquad -\infty<x<2.}
$$
This is the maximal connected interval containing the initial point. Direct <differentiation> gives $y'=1/(2-x)^2=y^2$; the pole at $x=2$ prevents continuation as a finite classical <solution of a differential equation> through that point.