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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.

In one period, sine equals $-\sqrt{3}/2$ at $4\pi /3$ and $5\pi /3$, and lies below that value between them. Since sine is a periodic function with period $2\pi$, all solutions are:The endpoints are included because the inequality is non-strict.