Question

Solve:

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

Answer 1

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

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

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

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

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

Let $y^2-1=t$

$\Rightarrow 2ydy=dt$ $ydy=dt/2$

and $\int \frac{\sqrt{x^2-a^2}}{x}dx=\sqrt{x^2-a^2}-a{\sec }^{-1}\left|\frac{|x|}{|a|}\right|+c$

So: $\int \frac{1}{t^{1/2}}\frac{dt}{2}=-\sqrt{x^2-1}+1\cdot {\sec }^{-1}|x|+c$

$=\frac{1}{2}\times \left(\frac{2}{3}\right)t^{\frac{1}{2}}=-\sqrt{x^2-1}+{\sec }^{-1}|x|+c$

$=\sqrt{y^2-1}=-\sqrt{x^2-1}+{\sec }^{-1}|x|+c$

$\Rightarrow \sqrt{y^2-1}+\sqrt{x^2-1}={\sec }^{-1}|x|+c$.