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Find the solution of $$\frac{xdx - ydy}{xdy - ydx} = \sqrt{\frac{1 - y^{2} + x^{2}}{x^{2} - y^{2}}}$$

I was able to bring it down to $$\frac{d(x^2-y^2)}{\sqrt{1+x^2-y^2}}=2\left(\frac{x.d(y/x)}{\sqrt{1-(y/x)^2}}\right)$$

Any help would be greatly appreciated. Thanks in advance! $$$$ Edit: An answer exists here, but I'm trying to solve the question $without$ using trigonometric or hyperbolic substitution (I was told it could be done without both, and that the solution was quite neat). $$$$ EDIT: Also the question that is mentioned as a possible duplicate is $\textbf{different}$ from mine.

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Ishan
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  • @Shakespeare I think the RHS should indeed have the extra $x$ on top. I'll just recheck my calculations. If the extra $x$ wasn't there, I'd be able to solve the question. – Ishan Sep 07 '16 at 12:04
  • You're right the $x$ should be there but you're missing a square root on the bottom, right? – ShakesBeer Sep 07 '16 at 12:12
  • Sorry, I forgot to add that. Thanks for notifying me. – Ishan Sep 07 '16 at 12:14
  • Try $d(\sqrt y / \sqrt x)$ instead of $d(y / x)$, remembering $d(\sqrt x)=dx/2 \sqrt x$ – ShakesBeer Sep 07 '16 at 12:15
  • @Shakespeare I'm sorry, I couldn't really work it out. Could you please show me how I should use $d(\sqrt y/\sqrt x)$? – Ishan Sep 07 '16 at 12:18
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    Dup of dup http://math.stackexchange.com/questions/1015021/solve-fracxdxydyxdy-ydx-sqrt-fraca2-x2-y2x2y2 – leonbloy Sep 07 '16 at 12:19
  • @Better World Polar coordinate conversion better? – Narasimham Sep 07 '16 at 12:23
  • @Shakespeare $$d(\sqrt y/\sqrt x)= \dfrac{xdy-ydx}{2x\sqrt x\sqrt y}$$ HOw should I proceed from here? – Ishan Sep 07 '16 at 12:23
  • @Narasimham Actually I was hoping for a solution without using Polar Coordinates. I've mentioned so in the edit. Could you please suggest another method? – Ishan Sep 07 '16 at 12:25
  • Better world, it was just an idea, it doesn't seem to actually work, sorry... – ShakesBeer Sep 07 '16 at 12:29
  • @leonbloy Kindly observe the difference between the question you have linked, and the question I have asked. – Ishan Sep 07 '16 at 12:29
  • Your question is exactly the same as this, which was considered already a dup http://math.stackexchange.com/questions/1906473/solve-fracxdx-ydyxdy-ydx-sqrt-frac1x2-y2x2-y2?noredirect=1&lq=1 – leonbloy Sep 07 '16 at 12:46
  • @leonbloy Please read the first of the 2 edits in the original question. Also, I was unable to suitably modify the first method in order to answer my own question. – Ishan Sep 07 '16 at 13:24
  • I have added an answer and I think it's what you have asked for. :-) – user90369 Sep 07 '16 at 14:55

3 Answers3

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From your reduction in comments mulpliply numerator & denominator of RHS by $\sqrt x$

$$ d(\sqrt y/\sqrt x)= \dfrac{xdy-ydx}{2x\sqrt x\sqrt y} =\dfrac{xdy-ydx}{2x^2 \sqrt {y/x}}$$

Let $(\sqrt y/\sqrt x) = u $ then

$$ 2 u du = d(u^2) = d (y/x) $$

$$ y/x = y/x + c $$

is bringing it back, puzzling...

Narasimham
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We can write the equation as follows:

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

It's $$ \frac{xdx-ydy}{x^2-y^2}=\frac{d\sqrt{x^2-y^2}}{\sqrt{x^2-y^2}}$$ and $$\frac{xdy-ydx}{x^2-y^2}=\frac{1}{2}d\ln\frac{x+y}{x-y}$$

$\,\displaystyle \sqrt{x^2-y^2}\,$ and $\,\displaystyle \ln\frac{x+y}{x-y}\,$ used as variables gives the wished result.

user90369
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  • @user90396 Thanks for your answer, it does indeed look like what I was hoping for! If you don't mind, could you please elaborate your last 2 steps? I couldn't quite understand what you had done. – Ishan Sep 07 '16 at 15:47
  • @user90306 I understood your method now Sir! Thanks very much! Lastly, could you please explain the motivation behind the first step$$\frac{xdx-ydy}{x^2-y^2}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}\frac{xdy-ydx}{x^2-y^2}$$? It's something that I couldn't have thought of myself, and so am very interested to know what made you think of it. – Ishan Sep 07 '16 at 15:53
  • I have only devided the equation by $x^2-y^2$. It was try and error. After getting the equation for $\frac{xdx-ydy}{x^2-y^2}$ the more difficult equation for $\frac{xdy-ydx}{x^2-y^2}$ could be determined. – user90369 Sep 07 '16 at 16:08
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The change of variables (hyperbolic coordinates) leads to a very simple ODE of the separable kind :

enter image description here

JJacquelin
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