Solution of Differential equation $\frac{xdx-ydy}{xdy-ydx} = \sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$ 
Solution of Differential equation $$\frac{xdx-ydy}{xdy-ydx} = \sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$$

$\bf{My\; Try::}$ Let $x=r\sec \theta$ and $y=r\tan \theta\;,$ Then $x^2-y^2=r^2$ and $xdx-ydy=rdr$
and $$\frac{y}{x} = \frac{\tan \theta }{\sec \theta} = \sin \theta\Rightarrow \frac{xdy-ydx}{x^2} = \cos \theta d\theta$$
So $$xdy-ydx=x^2\cos \theta = r^2\sec^2 \theta\cdot \cos \theta d\theta = r^2\sec \theta d\theta$$
So $$\frac{rdr}{r^2\sec \theta d\theta} = \sqrt{\frac{1+r^2}{r^2}}\Rightarrow \int \frac{dr}{\sqrt{1+r^2}} = \int \sec \theta d\theta$$
So $$\ln\left|1+\sqrt{1+r^2}\right| = \ln \left|\sec \theta +\tan \theta\right|+\ln \mathcal{C}$$
So $$\left|1+\sqrt{1+x^2-y^2}\right|= \left|\mathcal{C}\cdot \left(\frac{x+y}{\sqrt{x^2-y^2}}\right)\right|$$
My Question is can we solve it any other shorter way, If yes then plz explain here
Thanks

A better change of variables (in hyperbolic system of coordinates) is :
$\begin{cases}
    x=r \cosh(t)\\
    y=r \sinh(t)\\
  \end{cases}$ 
$\begin{cases}
    dx=dr \cosh(t)+r \sinh(t) dt\\
    dy=dr \sinh(t)+r \cosh(t) dt\\
  \end{cases}$
Bringing them into :
$$\frac{xdx-ydy}{xdy-ydx} = \sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$$
leads to :
$$\frac{1}{r}\frac{dr}{dt}=\sqrt{\frac{1+r^2}{r^2}}$$
$$t=\pm\int \frac{dr}{\sqrt{1+r^2}}=\pm \sinh^{-1}(r)+c$$
$$r=\pm\sinh(t-c)$$
The solution on parametric form is : 
$\begin{cases}
    x=\pm\sinh(t-c) \cosh(t)\\
    y=\pm\sinh(t-c) \sinh(t)\\
  \end{cases}$
The elimination of $t$ from the parametric system leads to the implicit equation :
$$x^2-y^2-y\:\cosh(c)+x\:\sinh(c)=0$$
