Trouble finding the integral curves of a PDE I have to verify that $ C1=x/(x+y+z) $ and $ C2=y^2-z^2 $ are the integral curves of the PDE 
$ x∂z/∂x + z∂z/∂y = y $
I already verified that C2 is (using method of characteristic curves) by solving
$ ∂y/z=∂z/y $
but when i try solving
$ ∂x/x=∂z/sqrt(C2+z^2) $
the result is something that has nothing to do with C1, the result is 
$ C1=ln(x)-(arctan(z/(y^2-z^2))/sqrt(y^2-z^2) $
How can i get C1? I also tried solving $ ∂x/x=∂y/sqrt(y^2-C2) $ but the result is something with exponentials
 A: $$x\frac{\partial z}{\partial x}+z\frac{\partial z}{\partial y}=y$$
System of characteristic ODEs (Charpit-Lagrange) :
$$\frac{dx}{x}=\frac{dy}{z}=\frac{dz}{y}$$
A first characteristic equation comes from 
$$\frac{dx}{x}=\frac{dy}{z}=\frac{dz}{y}=\frac{dx+dy+dz}{x+y+z}=\frac{d(x+y+z)}{(x+y+z)} \tag {Note below}$$
Solving $\frac{dx}{x}=\frac{d(x+y+z)}{(x+y+z)}$leads to $\ln|x|=\ln|x+y+z|+$constant.
$$\frac{x}{x+y+z}=C_1$$
A second characteristic equation comes from solving $\frac{dy}{z}=\frac{dz}{y}$ :
$$y^2-z^2=C_2$$
NOTE : Remember a wellknown property of fractions :
$$\frac{A}{B}=\frac{C}{D}=\frac{E}{F}\quad\implies\quad \frac{A}{B}=\frac{C}{D}=\frac{E}{F}=\frac{c_1A+c_2C+c_3E}{c_1B+c_2D+c_3F}$$
for arbitrary constants $c_1,c_2,c_3$ not all nul.
Another (but equivalent method) :
$$\frac{dx}{x}=\frac{dy}{z}=\frac{dz}{y}=dt \quad\implies\quad 
\begin{cases}
dx=x\:dt\\ dy=z\:dt\\ dz=y\:dt
\end{cases}  \quad\implies\quad dx+dy+dz=xdt+zdt+ydt$$
$$d(x+y+z)=(x+y+z)dt=(x+y+z)\frac{dx}{x}  \quad\implies\quad \frac{d(x+y+z)}{(x+y+z)}=\frac{dx}{x}$$
