How to find the second general solution of this equation? $\frac{dx}{y(x+y)+az} = \frac{dy}{x(x+y)-az} = \frac{dz}{z(x+y)}$
By comparing the first two ratios (adding them), I found one solution. How do find the second solution?
 A: $$\frac{dx}{y(x+y)+az} = \frac{dy}{x(x+y)-az} = \frac{dz}{z(x+y)}$$
$\frac{dx}{y(x+y)+az} = \frac{dy}{x(x+y)-az} =\frac{dx+dy}{(x+y)^2}$
$$\frac{dx+dy}{(x+y)^2}= \frac{dz}{z(x+y)}$$
$$\frac{dx+dy}{x+y}= \frac{dz}{z}$$
$$\boxed{\frac{z}{x+y}=c_1}$$
$\frac{dx}{y(x+y)+az} = \frac{dy}{x(x+y)-az}=\frac{dx}{(y+ac_1)(x+y)} = \frac{dy}{(x-ac_1)(x+y)}$
$$\frac{dx}{y+ac_1} = \frac{dy}{x-ac_1}$$
$$(x-ac_1)dx-(y+ac_1)dy=0$$
$$\frac12(x^2-y^2)-ac_1(x+y)=c_2$$
$$\boxed{\frac12(x^2-y^2)-az=c_2}$$
The PDE is certainly 
$$(y(x+y)+az)\frac{\partial z}{\partial x}+(x(x+y)-az)\frac{\partial z}{\partial y}=(x+y)z$$
So, the general solution expressed on implicit form $c_1=F(c_2)$ is :
$$\boxed{z=(x+y)\:F\left(\frac12(x^2-y^2)-az \right)}$$
$F$ is an arbitrary function, to be determined according to some boundary condition (presently not specified in the wording of the question).
AN ALTERNATIVE WAY TO SOLVE THE PDE :
Change of function 
$$z=(x+y)u\quad\implies\quad \begin{cases} z_x=u+(x+y)u_x \\z_y=u+(x+y)u_y \end{cases}$$
$(y(x+y)+a(x+y)u)(u+(x+y)u_x) +(x(x+y)-a(x+y)u)(u+(x+y)u_y)=(x+y)^2u$
After simplification :
$$(y+au)\frac{\partial u}{\partial x}+(x-au)\frac{\partial u}{\partial y}=0$$
Solving with method of characteristics leads to :
$$u=F\left(\frac12(x^2-y^2)-a(x+y)u \right)$$
