Question:
Find the general solution of the differential equation $\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0$
Solution:
$\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0$
$\Rightarrow \frac{d y}{d x}=-\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$
$\Rightarrow \frac{d y}{\sqrt{1-y^{2}}}=\frac{-d x}{\sqrt{1-x^{2}}}$
Integrating both sides, we get:
$\sin ^{-1} y=-\sin ^{-1} x+\mathrm{C}$
$\Rightarrow \sin ^{-1} x+\sin ^{-1} y=\mathrm{C}$