Solution
= Solution
Again $C=0$. Since the initial sine is nonzero, the local branch must satisfy $\sin y=x$. Selecting the inverse-sine branch through $y=\pi/6$ gives
$$
\boxed{y(x)=\arcsin x,\qquad -1<x<1}.
$$
This passes through the specified point and has $y'=1/\sqrt{1-x^2}$, which verifies the original equation on this interval.