Example 5 y′ = 2x√___

Solve

y^{'} = 2 x \sqrt{1 - y^{2}}

Integrating gives

\begin{aligned} \int \frac{d y}{\sqrt{1 - y^{2}}} & = \int 2 x d x \sqrt{1 - y^{2}} \neq 0 \\ \arcsin (y) & = x^{2} + c \\ y & = \sin (x^{2} + c) \end{aligned}

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{aligned} y & = \sin (x^{2} + c) \\ y & = 1 \\ y & = - 1 \end{aligned}