Question

Solve:
$\frac{dy}{dx}+\sqrt{\frac{1-y^2}{1-x^2}}=0$

Answer 1

$\frac{dy}{\sqrt{1-y^2}}+\frac{dx}{\sqrt{1-x^2}}=0$

Integrating we have,

${\sin }^{-1}y+{\sin }^{-1}x=c$ [Using direct formula and $c$ being integrating constant]

or, ${\sin }^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})=c$

or, $(x\sqrt{1-y^2}+y\sqrt{1-x^2})=\sin c$