Question

Solve the differential equation $\frac{dy}{dx}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$.

Answer 1

$\to \frac{dy}{dx}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$

$\int \frac{dy}{\sqrt{1-y^2}}=\int \frac{dy}{\sqrt{1-x^2}}$

${\sin }^{-1}y={\sin }^{-1}x+c$

$y=\sin ({\sin }^{-1}x+c)$