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I have come accross a problem which leads to the following ODE:

$\frac{dy}{dx}= \frac{y}{x}-\frac{1}{h}\frac{\sqrt{x^2+y^2}}{x}$,

where $h>0$ is a parameter. I am not able to solve it, however maple gives an implicit solution:

$\frac{x^{1/h}y}{x}+\frac{x^{1/h}\sqrt{x^2+y^2}}{x}=$constant.

I want to understand, how to $\textit{find}$ this implicit solution. (Of course one can differentiate it and find back the ODE.)

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    As you can see from the [tag-wiki](http://math.stackexchange.com/tags/differential-equations/info), the tag [tag:differential-equations] is intended for ODEs. There's no need to create a new tag.2012-10-28

1 Answers 1

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The important observation is that th equation is homogeneous. Indeed

$\eqalign{ & \frac{{dy}}{{dx}} = \frac{y}{x} - \frac{1}{h}\frac{{\sqrt {{x^2} + {y^2}} }}{x} \cr & \frac{{dy}}{{dx}} = \frac{y}{x} - \frac{1}{h}\sqrt {\frac{{{x^2} + {y^2}}}{{{x^2}}}} \cr & \frac{{dy}}{{dx}} = \frac{y}{x} - \frac{1}{h}\sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}} \cr} $

Thus, let $y=vx$. Then $y'=v'x+v$; and we get

$\eqalign{ & \frac{{dv}}{{dx}}x + v = v - \frac{1}{h}\sqrt {1 + {v^2}} \cr & \frac{{dv}}{{dx}}x = - \frac{1}{h}\sqrt {1 + {v^2}} \cr & \frac{{dv}}{{\sqrt {1 + {v^2}} }} = - \frac{{dx}}{x}\frac{1}{h} \cr} $

Can you move on? The complete solution would be:

$\begin{eqnarray*} {\sinh ^{ - 1}}v = - \frac{1}{h}\left( {\log x + C} \right)\\\log \left( {v + \sqrt {1 + {v^2}} } \right) = \log \frac{1}{{\root h \of x }} - \frac{C}{h}\\ \log \left( {\frac{y}{x} + \sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}} } \right) = \log \frac{1}{{\root h \of x }} - \frac{C}{h} \\\frac{y}{x} + \sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}} = \frac{k}{{\root h \of x }}\\y + \sqrt {{x^2} + {y^2}} = k{x^{1 - \frac{1}{h}}} \end{eqnarray*}$

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    Thank you! I thought about homogenity, but should have computed a bit. Now all is Ok.2012-10-26