Question

The solution of $x\sqrt{1-y^2}dx+y\sqrt{1-x^2}dy=0$ is:

• A. $si{n}^{-1}x+si{n}^{-1}y=c$
• B. $\sqrt{1-x^2}+\sqrt{1-y^2}=c$
• C. $\sqrt{1-x^2}\sqrt{1-y^2}=c$
• D. $\sqrt{\frac{1-x^2}{1-y^2}}=c$

Answer 1

The correct option is B $\sqrt{1-x^2}+\sqrt{1-y^2}=c$

$\int -\frac{x}{\sqrt{1-x^2}}dx=\int \frac{-y}{\sqrt{1-y^2}}dy+c$
$\Rightarrow -\sqrt{1-x^2}=\sqrt{1-y^2}+c$ $\therefore \frac{d}{dx}\sqrt{1-x^2}=\frac{-x}{\sqrt{1-x^2}}$
$\Rightarrow \sqrt{1-x^2}+\sqrt{1-y^2}=c$