To find the integral surface of given differential equation. Find the integral surface of the differential equation $(x-y)p+(y-x-z)q=z$ passing through the circle C: $z=1, x^2+y^2=1$
Clearly the Lagrange's auxillary equations are
$\frac{dx}{P}     = \frac{dy}{Q}. =\frac{dz}{R}$
Where P=$(x-y)$ ,Q=$y-x-z$ & R=$z$
on comparing the given P.D.E with 
the general quasilinear equation 
P(x,y,z) p+Q(x,y,z)q=R(x,y,z)
I obtained two solutions
$x+y+z=c_1$
&
$\frac{x-y+z}{z^2}=c_2$
Then I substituted $x=s$ where S is a parameter.
So $y=\sqrt{1-s^2}$ & $z=1$.
I need to find a relation in $c_1$ &$c_2$
Then substitute for $c_1$ &$c_2$ to find the final integral surface passing through given circle.
How can I proceed now?
 A: Your calculus is correct. The characteristic equations are :
$$\begin{cases}
x+y+z=c_1\\
\frac{x-y+z}{z^2}=c_2
\end{cases}$$
The general solution of the PDE, expressed on the form of implicit equation, is :
$$F(X,Y)=0 \quad \begin{cases}
X=x+y+z\\
Y=\frac{x-y+z}{z^2}
\end{cases}$$
where $F(X,Y)$ is any differentiable equation of two variables.This function has to be determined according to the conditions :
First condition : $z=1
\begin{cases}
X=x+y+1\\
Y=x-y+1
\end{cases}
\quad\to\quad 
\begin{cases}
x=\frac{X+Y}{2}-1\\
y=\frac{X-Y}{2}
\end{cases}$
Second condition : $x^2+y^2=1 \quad\to\quad \left(\frac{X+Y}{2}-1\right)^2+\left(\frac{X-Y}{2}\right)^2=1$
After simplification : $X(X-2)+Y(Y-2)=0$ which determines the function 
$$F(X,Y)=X(X-2)+Y(Y-2)$$
Thus, with 
$\begin{cases}
X=x+y+z\\
Y=\frac{x-y+z}{z^2}
\end{cases}$
 the particular solution according to the specified condition is :
$$(x+y+z)(x+y+z-2)+\frac{x-y+z}{z^2}\left(\frac{x-y+z}{z^2}-2\right)=0$$
This is the equation of the surface passing through the specified circle.
