Question

$\frac{dy}{dx}+\frac{\sqrt{(x^2-1)(y^2-1)}}{xy}=0$
Solve the above equation:

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

Answer 1

Answer: (A), (B)

The correct options are
A $\sqrt{x^2-1}-{\sec }^{-1}x+\sqrt{y^2-1}=c$
B $\sqrt{x^2-1}-{\sec }^{-1}x+\sqrt{y^2-1}=-c$

$\frac{dy}{dx}=\frac{\sqrt{x^2-1}\sqrt{y^2-1}}{xy}\Rightarrow \frac{y}{\sqrt{y^2-1}}dy=\frac{\sqrt{x^2-1}}{x}dx$
$\Rightarrow \int \frac{y}{\sqrt{y^2-1}}dy=\int \frac{\sqrt{x^2-1}}{x}dx$
$\Rightarrow -\sqrt{y^2-1}={\tan }^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)+\sqrt{x^2-1}+c\Rightarrow \sqrt{x^2-1}+\sqrt{y^2-1}-{\sec }^{-1}x=c$