Question

Find the general solution of differential equation $\frac{dy}{dx}+\sqrt{\frac{1-y^2}{1-x^2}}=0$; $x\ne 1$.

Answer 1

Given differential equation can be written as,

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

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

Now integrate both sides,

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

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