Exercise :
Solve the PDE $$x^2u_x + y^2u_v + z(x+y)u_z = 0$$
Attempt :
We have to solve the problem :
$$\frac{dx}{x^2} = \frac{dy}{y^2} = \frac{dz}{z(x+y)}$$
Choosing the first two fractions, we yield :
$$\frac{dx}{x^2} = \frac{dy}{y^2} \Rightarrow \int \frac{dx}{x^2} = \int \frac{dy}{y^2} \Leftrightarrow -\frac{1}{x} = \frac{1}{y} + c $$ $$\implies $$ $$u_1 = \frac{1}{x} - \frac{1}{y}$$
Now, choosing the first and the last fraction, we yield :
$$\frac{dx}{x^2} = \frac{dz}{z(x+y)} \Rightarrow \int z(x+y)dx = \int x^2dz \Leftrightarrow z\frac{x^2}{2} + zxy = x^2z + c' $$
but this is a different result for $u_2$ and eventually the general solution $u(x,y,z) = F(u_1,u_2)$ than Wolfram Alpha calculates here. Why is this the case and where am I wrong with my solution ?
For the first equation
$$\frac{dx}{x^2} = \frac{dy}{y^2} \Rightarrow \int \frac{dx}{x^2} = \int \frac{dy}{y^2} \Leftrightarrow -\frac{1}{x} = -\frac{1}{y} + C$$
$$ \implies K=\frac 1x- \frac 1y $$ For the last equation use this trick @Rebellos
$$\frac{dx}{x^2} = \frac{dy}{y^2} = \frac{dz}{z(x+y)}$$ $$\frac{dx-dy}{x^2-y^2} = \frac{dz}{z(x+y)}$$ $$\frac{d(x-y)}{x-y} = \frac{dz}{z}$$
I let you finish this last equation