Answer
$y=\sin (x^2+c)$
Work Step by Step
Given: $\dfrac{dy}{dx}=2x \sqrt {1-y^2}$
Re-arrange the given equation and integrate as follows:.
Then $\int(2x)dx= \int \dfrac{dy}{\sqrt {1-y^2}}$....(1)
As we know that $\int x^n dx=\dfrac{x^{n+1}}{n+1}+c$
Equation (1) becomes: $ \sin^{-1} y=x^2+c$
Now, consider taking $\sin$ of both sides. Then $\sin (\sin^{-1} y)=\sin (x^2+c)$
Hence, $y=\sin (x^2+c)$
