Given this problem
$$ \begin{cases} 2u_x - 5u_y + 3u_z=0\\ u(0,0,z)=z+3 \end{cases} $$
The general solution is:
$$u(x,y,z)=\Phi(-5x-2y,3y+5z)$$
Regarding $u(0,0,z)=z+3$, I saw that:
$$u(0,0,z)=\Phi(0,5z)=z+3$$
I tried to replace $5z$ expression with $3y+5z$, obtaining $$\frac{1}{5}(3y+5z)+3$$
but... how to handle the $-5x-2y$ formula?
Thanks
You can't because only one condition such as $\quad u(0,0,z)=z+3\quad$ is not sufficient to determine the function $\Phi$. $$u(x,y,z)=\Phi(-5x-2y,3y+5z)\quad\text{is correct}$$ $$u(x,y,z)=\Phi(X,Y)\quad\text{where}\quad\begin{cases}X=-5x-2y\\Y=3y+5z\end{cases}$$
With condition $\quad u(0,0,z)=z+3 =\Phi(X,Y)\quad\text{where}\quad\begin{cases}X=0\\Y=5z\end{cases}$
$$\Phi(0,Y)=\frac{1}{5}Y+3$$ One cannot determine $\Phi(X,Y)$ since this function is known only for $X=0$, not for all values of $X$.
Of course, this restrict the infinite set of functions $\Phi(X,Y)$ to subsets of functions but still remaining infinite.
$\frac{1}{5}Y+3=\frac{1}{5}(3y+5z)+3=\frac{3}{5}y+z+3$
For example and not exclusively: $$u(x,y,z)=\frac{3}{5}y+z+3+F\left(-5x-2y \right)$$ where $F$ is an arbitrary function such as $F(0)=0$.
Another example : $$u(x,y,z)=\left(\frac{3}{5}y+z+3\right)G(-5x-2y)$$ where $G$ is an arbitrary function such as $G(0)=1$.