$(x-y)p+(y-x-z)q=z$
Find the integral surface which the curves it passes are $z=1$ and $x^2+y^2=1$
Here is my try.
$$\frac{dx}{x-y}=\frac{dy}{y-x-z}=\frac{dz}{z}$$
So we have
$$\frac{dx+dy}{-z}=\frac{dz}{z}$$
We conclude $u=x+y+z=c_{1}$. How do we find other characteristic v?
Hint:
For the general solution,
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dz}{dt}=z$ , letting $z(0)=1$ , we have $z=e^t$
$\therefore\begin{cases}\dfrac{dx}{dt}=x-y~..........(1)\\\dfrac{dy}{dt}=y-x-e^t~......(2)\end{cases}$
Put $(1)$ into $(2)$ :
$\dfrac{dy}{dt}=-\dfrac{dx}{dt}-e^t$
$\dfrac{dx}{dt}+\dfrac{dy}{dt}=-e^t$
$\dfrac{d(x+y)}{dt}=-e^t$
$x+y=-e^t+C_1~......(3)$
Put $(3)$ into $(1)$ :
$\dfrac{dx}{dt}=2x+e^t-C_1$
$\dfrac{dx}{dt}-2x=e^t-C_1$
I.F. $=e^{\int-2~dt}=e^{-2t}$
$\therefore\dfrac{d(xe^{-2t})}{dt}=e^{-t}-C_1e^{-2t}$
$xe^{-2t}=-e^{-t}+\dfrac{C_1e^{-2t}}{2}+C_2$
$x=\dfrac{C_1}{2}+C_2e^{2t}-e^t$
$\therefore\begin{cases}x=\dfrac{C_1}{2}+C_2e^{2t}-e^t\\y=\dfrac{C_1}{2}-C_2e^{2t}\end{cases}$
$x(0)=x_0$ , $y(0)=f(x_0)$ :
$\begin{cases}\dfrac{C_1}{2}+C_2-1=x_0\\\dfrac{C_1}{2}-C_2=f(x_0)\end{cases}$
$\begin{cases}C_1=x_0+f(x_0)+1\\C_2=\dfrac{x_0-f(x_0)+1}{2}\end{cases}$
$\therefore\begin{cases}x=\dfrac{x_0+f(x_0)+1}{2}+\dfrac{(x_0-f(x_0)+1)e^{2t}}{2}-e^t\\y=\dfrac{x_0+f(x_0)+1}{2}-\dfrac{(x_0-f(x_0)+1)e^{2t}}{2}\end{cases}$
$\begin{cases}x+y=x_0+f(x_0)+1-e^t\\x-y=(x_0-f(x_0)+1)e^{2t}-e^t\end{cases}$
$\begin{cases}x_0+f(x_0)=x+y+e^t-1\\x_0-f(x_0)=(x-y)e^{-2t}+e^{-t}-1\end{cases}$
$\therefore\begin{cases}x_0=\dfrac{x+y+(x-y)e^{-2t}+e^t+e^{-t}-2}{2}=\dfrac{x+y+\dfrac{x-y}{z^2}+z+\dfrac{1}{z}-2}{2}=\dfrac{z^3+(x+y-2)z^2+z+x-y}{2z^2}\\f(x_0)=\dfrac{x+y-(x-y)e^{-2t}+e^t-e^{-t}}{2}=\dfrac{x+y-\dfrac{x-y}{z^2}+z-\dfrac{1}{z}}{2}=\dfrac{z^3+(x+y)z^2-z-x+y}{2z^2}\end{cases}$
Hence $\dfrac{z^3+(x+y)z^2-z-x+y}{2z^2}=f\biggl(\dfrac{z^3+(x+y-2)z^2+z+x-y}{2z^2}\biggr)$