Answer
\[y=\sin (C+\ln |x|)\]
Work Step by Step
\[xy'=(1-y^2)^{\frac{1}{2}}\;\;\;...(1)\]
\[x\frac{dy}{dx}=\sqrt{1-y^2}\]
Separating variables,
\[\frac{dy}{\sqrt{1-y^2}}=\frac{dx}{x}\]
Integrating,
\[\int\frac{dy}{\sqrt{1-y^2}}=\int\frac{dx}{x}+C\]
Where $C$ is constant of integration
$$ \sin^{-1} y=\ln |x|+C$$
\[y=\sin (C+\ln |x|)\]
Hence general solution of (1) is \[y=\sin (C+\ln |x|).\]
