$$w\partial_xl + x\partial_wl+(x+w)\partial_zl = 0;$$ $$ l = l(x, w) , \quad x + w > 0.$$
I was solving this question but I got stuck. After dividing by $x+y $, I get that $s = z$ and had $$ C_1=\tfrac12 x^2+zx-yz ,$$ $$C_2=\tfrac12 y^2 +yz-zx $$ but when verifying, I didn't get the right answer so can someone help me with this.
$$y\partial_xu + x\partial_yu+(x+y)\partial_zu = 0;$$ Solving with method of characteristics:
Since $x+y \ne 0$ : $$\frac {dx}{y}=\frac {dy}{x}=\frac {dz}{x+y}=\frac {du} 0$$ The first integral curve is easy to find: $$du=0 \implies c_0=u$$ $$ \frac {d(x+y)}{x+y}=\frac {dz}{x+y}$$ $$\implies c_1=x+y-z$$ Is an integral curve. $$\frac {dx}{y}=\frac {dy}{x}$$ $$xdx=ydy $$ $$c_2=x^2-y^2$$ Is another integral curve.
So that the integral curves are: $$ \begin{cases} c_0=u \\ c_1=x+y-z \\ c_2=x^2-y^2 \\ \end{cases} $$ Finally: $$\boxed {u(x,y,z)=F(x+y-z,x^2-y^2)}$$