Finding an Integral Surface Consider finding the integral surface of
$$x^2 p + xy q = xyz-2y^2$$
which passes through the line $x=y e^y$ in the $z=0$ plane.
Attempt
In Lagrange's subsidiary form
   $$\frac{dx}{x^2}=\frac{dy}{xy}=\frac{dz}{(xyz-2y^2)}$$
Firstly consider
   $$\frac{dx}{x^2}=\frac{dy}{xy}$$ 
One can trivially show that
   $$a = \frac{x}{y}$$
where $a$ is an arbitrary constant. Now, consider
   $$\frac{dx}{x^2}=\frac{dz}{(xyz-2y^2)}$$
which may be written as
   $$\frac{dz}{dx}=\frac{(xyz-2y^2)}{x^2} \equiv \frac{z}{a}-\frac{2}{a^2}$$
having used $a=x/y$ from before. (After this point I am unsure of my working...) 
   $$\frac{dz}{dx}=\frac{az-2}{a^2} \implies\frac{dz}{(az-2)}=\frac{dx}{a^2}$$


*

*As $a$ is a function of $a(x,y)$, albeit an arbitrary constant, is my solution above sensical or have a made a mistake?

*I understand that to find the integral surface the general solution is of the form $F(a,b)$ where has so far been determined to be $a=x/y$. How can I find this over arbitrary constant $b$?  
 A: Your calculus is correct. Don't worry about $a(x,y)$ which should be true outside the characteristic curves, but is constant on the characteristic curves. Thus it is legitim to integrate $\frac{dz}{dx}=\frac{z}{a}-\frac{2}{a^2}$ with $a=$ constant.
$$z=\frac{2}{a}+be^{x/a}$$
The second family of characteristic curves is :
$$e^{-x/a}(z-\frac{2}{a})=b$$
General solution of the PDE on the form of implicit equation $F(a,b)=0$ with any function $F$ of two variables, or equivalently  $b=\Phi(a)$ with any function $\phi$ of one variable :
$$b=e^{-x/a}(z-\frac{2}{a})=\Phi(\frac{x}{y})=e^{-y}(z-\frac{2y}{x})$$
$$z(x,y)=\frac{2y}{x}+e^y \Phi(\frac{x}{y})$$
$\Phi$ is an arbitrary function to be determined according to the boundary condition.
CONDITION : $x=ye^y$ on the plane $z=0$ .
$$\quad z(ye^y,y)=0=\frac{2y}{ye^y}+e^y \Phi(\frac{ye^y}{y})$$
$$\Phi(e^y)=-2 e^{-2y}=-2(e^{y})^{-2}$$
So, the function $\Phi$ is determined : $\Phi(X)=-2X^{-2}$ . We put it into the above general solution where $X=\frac{x}{y}$
$$z(x,y)=\frac{2y}{x}-2e^y \frac{y^2}{x^2}$$
A: Carry out the second integration, 
$$
az-2=be^{x/a}=be^y.
$$
Then use $b=\phi(a)$ or a similar dependence and insert the initial condition
$$
-2 = \phi(e^y)e^{y}\implies \phi(t)=-\frac{2}t.
$$
