How would this differential equation be solved? 
How would this differential equation be solved?
  $$y{\partial z\over \partial x}+z{\partial z\over \partial y}={y \over x}$$

I was taught to solve them like : $${dx \over y}={dy \over z}={dz \over {y \over x}}$$ Then find the constants $c_1$ and $c_2$ and the answer being $F(c_1,c_2)$ but I run into trouble with this one because i can't see a useful technique to get at least one $c$...
 A: From the hint you gave the answer is as follows :
$z = 1$ + $\frac{(y^2 - 2C)}{2x} $
Actually $dx = xdz$ 
gives
 $lnx = Be^z$
And then 
$x\frac{dz}{y}$ = $\frac{dy}{z}$ gives the following 
$y^2= 2Bze^z - 2Be^z + C$
 Which finally becomes -
 $ z = 1 + \frac{(y^2 - 2C)}{2x}.$
 Does it help ?
A: Here I provide answer to the second part
$\frac{dy}{dx}$= $2e^x -y$
$y +\frac{dy}{dx} = 2e^x$
$\frac{dy}{dx}+\frac{d^2y}{dx^2} = 2e^x$
$\frac{d^2y}{dx^2}=y$
$2\frac{dy}{dx}\frac{d^2y}{dx^2}=2\frac{dy}{dx}y $
$(\frac{dy}{dx})^2 = y^2 +c^2$
$\frac{dy}{\sqrt{{y^2}+c^2}}= dx$
$x= log |y+\sqrt{y^2 +c^2}|+A$
$$ Where A = log \frac{1}{2}$$
A: $y\dfrac{\partial z}{\partial x}+z\dfrac{\partial z}{\partial y}=\dfrac{y}{x}$
$x\dfrac{\partial z}{\partial x}+\dfrac{xz}{y}\dfrac{\partial z}{\partial y}=1$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dz}{dt}=1$ , letting $z(0)=0$ , we have $z=t$
$\dfrac{dx}{dt}=x$ , letting $x(0)=x_0$ , we have $x=x_0e^t=x_0e^z$
$\dfrac{dy}{dt}=\dfrac{xz}{y}=\dfrac{x_0te^t}{y}$ , we have $y^2=2x_0(t-1)e^t+f(x_0)=2x(z-1)+f(xe^{-z})$
