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


  • $\begingroup$ actuallly I have use $d(x^2-y^2) = d(r^2),$ Then $2xdx-2ydy=2rdr$ $\endgroup$
    – juantheron
    Jan 29 '16 at 7:44
  • $\begingroup$ This seems like a very elementary solution, why do you need a shorter way? (P.S., I saw that immediately as I commented...) $\endgroup$ Jan 29 '16 at 7:48

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 : $$x^2-y^2-y\:\cosh(c)+x\:\sinh(c)=0$$


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