Answer
$$ y = {\sin ^{ - 1}}x $$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }},\,\,\,\,\,y\left( 0 \right) = 0 \cr
& {\text{separating variables}} \cr
& dy = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr
& {\text{integrate both sides}} \cr
& y = \int {\frac{1}{{\sqrt {1 - {x^2}} }}} dx \cr
& y = {\sin ^{ - 1}}x + C\,\,\,\left( {\bf{1}} \right) \cr
& \cr
& {\text{use initial condition }}y\left( 0 \right) = 0 \cr
& 0 = {\sin ^{ - 1}}0 + C \cr
& C = 0 \cr
& \cr
& {\text{then}}{\text{, substitute }}C = 0{\text{ in }}\left( {\bf{1}} \right) \cr
& y = {\sin ^{ - 1}}x \cr} $$
