Solve $\frac{1}{x}z_x+\frac{1}{y}z_y=4$ $$\begin{cases}
\frac{1}{x}z_x+\frac{1}{y}z_y=4\\
z(1,y)=y^2-1.\\
\end{cases}$$
So we started with:
$$\frac{dx}{dt}=\frac{1}{x}\rightarrow x^2(t,s)=2t+f_1(s)$$
$$\frac{dy}{dt}=\frac{1}{y}\rightarrow y^2(t,s)=2t+f_2(s)$$
$$\frac{dz}{dt}=4\rightarrow z(t,s)=4t+f_3(s).$$
From $$z(1,y)=y^2-1$$ we get
$$x(0,s)=1,y(0,s)=s,z(0,s)=s^2-1.$$
How did we get $$x(0,s)=1,y(0,s)=s?$$
 A: This is follows by interpreting the initial condition $z(1,y)=y^2-1$ as $$z(x_0(s),y_0(s))=z_0(s).$$ Using $s=y$ as parameter of the underlying curve you get $x_0(s)=1$, $y_0(s)=s$, $z_0(s)=s^2-1$.

In the alternative formulation of the Lagrange equations they read as
$$
x\,dx=y\,dy=\frac{dz}4
$$
so that along characteristic curves $y^2-x^2=c_1$ and $z-2x^2=c_2=\phi(c_1)=\phi(y^2-x^2)$ gives the general solution as
$$
z(x,y)=2x^2+\phi(y^2-x^2).
$$
Directly inserting the initial condition then gives
$$
y^2-1=2+\phi(y^2-1)\implies \phi(u)=u-2\implies z(x,y)=x^2+y^2-2.
$$
A: You could also write the initial data curve before analyzing the characteristic equations.
The initial data curve is $\Gamma: \langle 1,s,s^2-1 \rangle$. The characteristic equations are 
$$\begin{cases} 
      \dfrac{dx}{dt}=\dfrac{1}{x}, &  x(0,s)=1 \\[1em]
      \dfrac{dy}{dt}=\dfrac{1}{y}, & y(0,s)=s \\[1em] 
      \dfrac{dz}{dt}=4, & z(0,s)=s^2-1
   \end{cases}
$$
For the first equation,
$$\frac{dx}{dt}=\dfrac{1}{x} ~\Rightarrow~ x^2(t,s) = 2t+c_1(s) $$
Plugging in the initial condition of $ x(0,s)=1 ~\Rightarrow~  c_1(s)=1$. Hence,
$$x^2(t,s)=2t+1$$
For the second equation,
$$\frac{dy}{dt}=\dfrac{1}{y} ~\Rightarrow~ y^2(t,s) = 2t+c_2(s) $$
Plugging in the initial condition of $ y(0,s)=s ~\Rightarrow~  c_2(s)=s^2$. Hence,
$$y^2(t,s) = 2t+s^2$$
For the last equation,
$$\frac{dz}{dt}=4 ~\Rightarrow~ dz = 4dt ~\Rightarrow~ z(t,s)=4t+c_3(s)$$
Plugging in the initial condition of $ z(0,s)=s^2-1 ~\Rightarrow~  c_3(s)=s^2-1$. Hence,
$$ z(t,s)=4t+s^2-1$$
Combining all three equations together, we see that
$$x^2(t,s)=2t+1 \implies t=\frac{x^2-1}{2}$$
$$y^2(t,s) = 2t+s^2 \implies s=\sqrt{y^2-2t}=\sqrt{y^2-x^2+1}$$
$$z(t,s)=4t+s^2-1$$
So, the solution is
$$z=u(x,y)=4t+s^2-1=4\Big(\frac{x^2-1}{2}\Big)+\Big(y^2-x^2+1\Big)-1=x^2+y^2-2$$
A: When writing a characteristic system of this PDE
$$\frac{dx}{dt} = \frac{1}{x},$$
$$\frac{dy}{dt} = \frac{1}{y},$$
$$\frac{dz}{dt} = 4,$$
then we can get constant functions on $t$ in the form 
$$I_{1}(x,y,z) = x^{2} - y^{2},$$
$$I_{2}(x,y,z) = 2x^{2} - z.$$
Now we are looking for an implicit function given by
$$\phi(I_{1},I_{2}) = \phi(x^{2} - y^{2}, 2x^{2} - z) = 0.$$
After imposing the initial condition we have
$$\phi(1 - y^{2}, 3 - y^{2}) = 0,$$
which implies that the possible form of the function $\phi(x,y,z)$ must be
$$\phi(x,y,z) = x^{2} + y^{2} - z - 2.$$ 
Thus, solution of the PDE with the given IC is
$$z(x,y) = x^{2} + y^{2} - 2.$$
