<s>[INST] Implicit solution to an ODE 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.)
 [/INST] 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:
$$\begin{eqnarray*} 
{\sinh ^{ - 1}}v =  - \frac{1}{h}\left( {\log x + C} \right)\\\log \left( {v + \sqrt {1 + {v^2}} } \right) = \log \frac{1}{{\root h \of x }} - \frac{C}{h}\\ \log \left( {\frac{y}{x} + \sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}} } \right) = \log \frac{1}{{\root h \of x }} - \frac{C}{h} \\\frac{y}{x} + \sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}}  = \frac{k}{{\root h \of x }}\\y + \sqrt {{x^2} + {y^2}}  = k{x^{1 - \frac{1}{h}}}
 \end{eqnarray*}$$
</s>