<s>[INST] 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.
 [/INST] 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.
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