Solve\[ y^{\prime }=2x\sqrt {1-y^{2}}\] Integrating gives\begin {align*} \int \frac {dy}{\sqrt {1-y^{2}}} & =\int 2xdx\qquad \sqrt {1-y^{2}}\neq 0\\ \arcsin \left ( y\right ) & =x^{2}+c\\ y & =\sin \left ( x^{2}+c\right ) \end {align*}
The singular solution is found by solving for \(y\) from \(\sqrt {1-y^{2}}=0\). This gives \(y^{2}=1\) or \(y=\pm 1\). Hence the solution is\begin {align*} y & =\sin \left ( x^{2}+c\right ) \\ y & =1\\ y & =-1 \end {align*}