# Solve this PDE $(y+z)u_x+(z+x)u_y+(x+y)u_z=0$

Solve $$(y+z)u_x+(z+x)u_y+(x+y)u_z=0$$

My method is to use characteristics.

$$dx/(y+z)=dy/(z+x)=dz/(x+y).$$ Solve this??

$d(x+y)/(x+y+2z)=dz/(x+y)$ which implies $$(x+y-2z)^2(x+y+z)=C_1.$$

And by symmetry, another independent first integral is $$(y+z-2x)^2(x+y+z)=C_2$$ and thus the solution of the PDE is $$u=\phi((x+y-2z)^2(x+y+z),(y+z-2x)^2(x+y+z)).$$Is this right?

Since you have $x'=y+z$, $y'=z+x$, and $z'=x+y$, you have the system $$\frac{d}{dt} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}$$
In general, $\vec{v}'=A\vec{v} \implies \vec{v}=\mathrm{e}^{At}\vec{v}_0$.
Supposing you can find $\mathrm{e}^{At}$ (I bet $A$ in this case is easily diagonalizable), then you have $\frac{du}{dt}=0\implies u(x(t;\vec{v}_0),y(t;\vec{v}_0),z(t;\vec{v}_0)) = c$. It's difficult to say more without some side conditions, but at this point it's now a standard exercise in characteristics.