Transform the partial differential equation $(y-z)\frac{∂z}{∂x}+(y+z)\frac{dz}{dy}=0$ Transform the P.D.E.
$(y-z)\frac{\partial z}{\partial x}+(y+z)\frac{\partial z}{\partial y}=0$
so that the new equation contains $x$ as a new function, and $u=y-z, v=y+z$ are new independent variables.
 A: $$u=y-z\quad\implies\quad du=dy-dz$$
$$v=y+z\quad\implies\quad dv=dy+dz$$
$$dx=\frac{\partial x(u,v)}{\partial u}du+\frac{\partial x(u,v)}{\partial v}dv$$
$$dx=\frac{\partial x(u,v)}{\partial u}(dy-dz)+\frac{\partial x(u,v)}{\partial v}(dy+dz)$$
$$dx=\left(\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}\right)dy+\left(-\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}\right)dz$$
$$dz=\frac{1}{-\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}dx-\frac{\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}{-\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}dy\quad\implies$$
$$\implies\quad\begin{cases}
\frac{\partial z(x,y)}{\partial x}=\frac{1}{-\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}\\
\frac{\partial z(x,y)}{\partial y}=-\frac{\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}{-\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}
\end{cases}$$
We put them into the original PDE where $y-z=u$ and $y+z=v$ :
$$(y-z)\frac{\partial z}{\partial x}+(y+z)\frac{\partial z}{\partial y}=0$$
$$u\frac{1}{-\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}-v\frac{\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}{-\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}=0$$
$$u-v\left(\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}\right)=0$$
$$\boxed{\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}=\frac{u}{v}}$$
This is the transformed PDE we were looking for.
This PDE can be solved thanks to the method of characteristics. The result is :
$$\boxed{x(u,v)=v+(u-v)\ln|v|+F(u-v)}$$
$F$ is an arbitrary function until some boundary condition be specified.
Comming back to the original variables :
$$x=y+z(x,y)-2z(x,y)\ln|y+z(x,y)|+F\big(-2z(x,y)\big)$$
or equivalenttly
$$\boxed{z\ln|y+z|=\frac{y-x}{2}+G(z)}$$
$G$ is an arbitrary function until some boundary condition be specified.
The last equation cannot be solve for $z$ in terms of standard functions. Thus one have to be satisfied for the solution expressed on the form of implicit equation.
NOTE : The original PDE is even easier to solve without the change of variables. The direct solving leads to the solution consistent with the above solution. This is a manner to check that the above result is correct.
