Solution of the first order partial differential equation . The integral surface of the first order partial differential equation $$2y(z-3)\frac{\partial z}{\partial x}+(2x-z)\frac{\partial z}{\partial y} = y(2x-3)$$ passing through the curve $x^2+y^2=2x, z = 0$ is


*

*$x^2+y^2-z^2-2x+4z=0$ 

*$x^2+y^2-z^2-2x+8z=0$ 

*$x^2+y^2+z^2-2x+16z=0$ 

*$x^2+y^2+z^2-2x+8z=0$
My effort:
I find the mulipliers $x, 3y, -z$ and get the solution $x^2+3y^2-z^2=c_1$. How to proceed further? Please help.
 A: Four equations are proposed and it is asked to find which one satisfies both properties :


*

*First : $x^2+y^2=2z$ at $z=0$

*Second : Is solution of the PDE.
HINT :
By inspection, it is obvious that the four equations satisfy the first property.
So, the question is to find which one of the four proposed equations satisfy the PDE.
Why not simply putting the equations, one after the other, into the PDE and see if it agrees or not ?
I suppose that you can do it. 
Nevertheless, the next calculus might help you.
For example, with the equation $\quad x^2+y^2+z^2-2x+4z=0 \quad$. (This is not one of the four proposed equations, but another example).
$$2xdx+2ydy+2zdz-2dx+4dz=0$$
$$dz=\frac{-2x+2}{2z+4}dx+\frac{-2y}{2z+4}dy$$
$$\frac{\partial z}{\partial x}=\frac{-2x+2}{2z+4}\quad;\quad \frac{\partial z}{\partial y}=\frac{-2y}{2z+4}$$
$$2y(z-3)\frac{-2x+2}{2z+4} +(2x-z)\frac{-2y}{2z+4} - y(2x-3) =$$
$$=-4y\frac{xz+3}{z+2}$$
This is not equal to $0$. Thus the equation $x^2+y^2+z^2-2x+4z=0$ is not solution of the PDE.
No need to solve the PDE with the method of characteristics to answer to the question. Thanks to this simple method of checking, you will easily find that equation 1 is the right answer.
A: All surfaces passing through the curve $x^2+y^2=2x, z = 0$. Bat only 
first surface $$x^2+y^2-z^2-2x+4z=0\tag{1}$$ is solution of pde. We check it
Let $z=z(x,y)$. We differentiate equation $(1)$ respect $x$ and $y$:
$$2x-2zz'_x-2+4z'_x=0,\\
2y-2zz'_y+4z'_y=0.
$$
Solving this system we get
$$z'_x=\frac{x-1}{z-2},\quad z'_y=\frac{y}{z-2}$$
Next  we substitute this in to pde. We get
$$\frac{2 \left( x-1\right)  y\, \left( z-3\right) }{z-2}+\frac{y\, \left( 2 x-z\right) }{z-2}=\left( 2 x-3\right) \, y,$$
$$(2x-3)y=(2x-3)y.$$
Then $(1)$ is solution.
