Question

Solve:

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

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

Answer 1

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

Given differential eqn can be written as
$\frac{dy}{dx}=-\frac{\sqrt{(x^2-1)(y^2-1)}}{xy}$
$\Rightarrow \frac{y}{\sqrt{y^2-1}}dy=-\frac{\sqrt{x^2-1}}{x}dx$
Integrating both sides
$\int \frac{y}{\sqrt{y^2-1}}dy=-\int \frac{\sqrt{x^2-1}}{x}dx$
Put $y^2-1=t$
$\Rightarrow 2ydy=dt$
Also, put $\sqrt{x^2-1}=u$
$\Rightarrow x^2-1=u^2$
$\Rightarrow xdx=udu$
$\int \frac{1}{2\sqrt{t}}dt=-\int \frac{u}{1+u^2}udu$
$\Rightarrow \int \frac{1}{2\sqrt{t}}dt=-\int \frac{u^2}{1+u^2}du$
$\Rightarrow \sqrt{t}=-\int (1-\frac{1}{1+u^2})du$
$\Rightarrow \sqrt{y^2-1}=-u+{\tan }^{-1}u+C$
$\Rightarrow \sqrt{y^2-1}=-\sqrt{x^2-1}+{\tan }^{-1}\sqrt{x^2-1}+C$
$\Rightarrow \sqrt{y^2-1}=-\sqrt{x^2-1}+{\sec }^{-1}x+C$ (${\tan }^{-1}\sqrt{x^2-1}=z\Rightarrow \sqrt{x^2-1}=\tan z\Rightarrow \sec z=x$ )