The equation $ pq = px + qy $ has more than one complete integral. Show that the equation $ pq = px + qy $ has more than one complete integral.
Using Charpit Method I find $az = \frac{1}{2}(y + ax)^{2} + b$, where $a$ and $b$ are integrating constant. How could I find another? Is there any theorem to show for existence of more than one complete integral?
Please help.
 A: $$pq=xp+yq\quad ;\quad p=\frac{\partial z(x,y)}{\partial x}\quad ;\quad q=\frac{\partial z(x,y)}{\partial y}$$
If we look only for particular solutions there is no need for the characteristic method. Simple inspection draw to try a function on the form $z=f(xy)$ for example.
$$p=yf'\quad;\quad q=xf'\quad;\quad xy(f')^2=x(yf')+y(xf')=2xyf'\quad;\quad f'=2$$
$$\boxed{z(x,y)=2xy+c}$$
This is sufficient to show that there is more than one solution $az=\frac12(y+ax)^2+b$ .
Note that this solution $az=\frac12(y+ax)^2+b$ can be obtained directly in searching particular solutions on the form $z=f(\alpha y+\beta x)$.
With the Charpit-Lagrange method the system of characteristic ODEs is :
$$\frac{dx}{x-q}=\frac{dy}{y-p}=\frac{dp}{-p}=\frac{dq}{-q}=\frac{dz}{(x-q)p+(y-p)q}$$
Note that the last fraction is of no interest since this is the combination of the two first fractions and equivalent to the obvious $dz=p\,dx+q\,dy$.
No need to repeat here the well known method which for example is explained in full details in this video : https://www.youtube.com/watch?v=U51hN02LYrw
The solution $z=2xy+c$ is straightforward in observing that the system of ODEs is satisfied with $p=2y$ and $q=2x$:
$$\frac{dx}{x-(2x)}=\frac{dy}{y-(2y)}=\frac{d(2y)}{-(2y)}=\frac{d(2x)}{-(2x)}$$
$\frac{\partial z}{\partial x}=2y\implies z=2xy+g(y)\quad$ and $\quad\frac{\partial z}{\partial y}=2x\implies z=2xy+h(x)$
$g(y)=h(x)\implies =c \quad\implies z=2xy+c$
