Solve the Cauchy problem by the method of characteristic $pz+q=1$ with initial data $y=x, z=x/2$. Indicate the region where the solution is valid.
How to solve this problem.
Lagrange's auxiliary equation $$\frac{dx}{z}=\frac{dy}{1}=\frac{dz}{1}$$ Integrating $y-z=c_1$, $z^2-2x=c_2$.
Using $y=x, z=x/2$ i.e. $y=x=t, z=t/2$, the equations $y-z=c_1$, $z^2-2x=c_2$ give $t=2c_1$, $t^2-8t=4c_2$.
Eliminating $t$, we have $c_1^2-4c_1-c_2=0$.
If it is right, what will be the solution? Please indicate the region where the solution is valid. By putting $c_1$, $C_2$, is $(y-z)^2-4(y-z)-(z^2-2x)=0$ the solution?
$$\frac{dx}{z}=\frac{dy}{1}=\frac{dz}{1}\quad\text{is OK.}$$ OK for the characteristic equations $\quad y-z=c_1\quad$ and $\quad z^2-2x=c_2$.
The general solution $\Phi(c_1\:,\:c_2)=0$ is : $$\Phi(y-z\:,\:z^2-2x)=0$$ $Phi$ is an arbitrary function on two variables.
Or on the form of implicit equation : $$\boxed{z^2-2x=F(y-z)}$$ where $F$ is an arbitrary function (to be determined according to the specified condition).
Condition : $y=x, z=\frac12 x \quad\implies\quad (\frac12 x)^2-2x=F(x-\frac12 x)$
Let $X=\frac12 x$
$(\frac12 x)^2-2x=F(x-\frac12 x)=(X)^2-4X=F(X)$ $$F(X)=X^2-4X$$ The function $F(X)$ is determined. We put it into the above general solution where $X=y-z$. $$z^2-2x=(y-z)^2-4(y-z)$$ This is consistent with the solution that you found. Your solution is valid.
After simplification : $$z(x,y)=\frac{y^2-4y+2x}{2(y-2)}$$