Interpretation of solution of pde of form $f(u,v)=0$ Given PDE
$zz_x-zz_y=z^2+(x+y)^2$
I had solved problem and found that solution is 
$F(x+y,\log(x^2+y^2+z^2+2xy)-2x)=0$
I know how to solve Above type of PDE.
But problem is that I do not know what is above function means? 
Please can someone kindly give information about solution denoted by function $F(u(x,y),v(x,y))=0$ ?
Any Help will be appreciated.
 A: $$zz_x-zz_y=z^2+(x+y)^2$$
The Charpit-Lagrange system of equations is : https://en.wikipedia.org/wiki/Method_of_characteristics
$$\frac{dx}{z}=\frac{dy}{-z}=\frac{dz}{z^2+(x+y)^2}$$
A first characteristic equation comes from $\frac{dx}{z}=\frac{dy}{-z}$  which leads to :
$$x+y=c_1$$ 
A second characteristic equation comes from $\frac{dx}{z}=\frac{dz}{z^2+(x+y)^2}$ which leads to :
$\frac{dx}{z}=\frac{dz}{z^2+c_1^2}\quad;\quad 2dx=\frac{d(z^2)}{z^2+c_1^2}\quad;\quad 2x=\ln(z^2+c_1^2)+$constant.
$$\ln(z^2+c_1^2)-2x=c_2$$
$$\ln(z^2+(x+y)^2)-2x=c_2$$
$$\ln(x^2+y^2+z^2+2xy)-2x=c_2$$
The general solution of the PDE expressed on the form of implicite equation $F(c_1,c_2)=0$ is :
$$F\big((x+y)\:,\: (\ln(x^2+y^2+z^2+2xy)-2x)\big)=0$$
where $F$ is an arbitrary function of two variables.
This means that you can take the function $F$ as you like. 
For example if you take $F(X,Y)=X^3-Y^2$ you get $(x+y)^3-(\ln(x^2+y^2+z^2+2xy)-2x)^2=0$ which is a particular solution of the PDE expressed on the form of implicit equation.
Since they are an infinity of functions $F$ they are an infinity of solutions insofar no boundary condition is specified.
In order to determine the function $F$ some boundary conditions must be specified.
NOTE : 
Equivalently the solutions can be expressed on explicit form $c_2=\phi(c_1)$ :
$$\ln((x+y)^2+z^2)-2x=\phi(x+y)$$
where $\phi$ is an arbitrary function.
$$(x+y)^2+z^2=e^{\phi(x+y)+2x}=e^{2x}f(x+y)$$
where $f$ is an arbitrary function.
$$z=\pm\sqrt{e^{2x}f(x+y)-(x+y)^2}$$
Some boundary conditions must be specified in order to determine the function $f$.
