Question

If $\frac{dy}{dx}+\sqrt{\frac{1-y^2}{1-x^2}}=0$. Prove that, $x\sqrt{1-y^2}+y\sqrt{1-x^2}=A$ where A is constant.

Answer 1

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

$\Rightarrow {\sin }^{-1}y=-{\sin }^{-1}x+c$ [c is a constant]

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

$={\sin }^{-1}[x\sqrt{1-y^2}+y\sqrt{1-x^2}]=c$

$\Rightarrow x\sqrt{1-y^2}+y\sqrt{1-x^2}=\sin c=A$

Where A is a constant.