QUESTİON:
x(y^2+z)z_x - y(x^2+z)z_y =(x^2-y^2).z
thanks for your helps dear friends.
The characteristic equations are
$\displaystyle \frac{dx}{x(y^2+z)} = \frac{dy}{-y(x^2+z)} = \frac{dz}{(x^2- y^2)z}
$
from which we can deduce
$\displaystyle \frac{dx}{x} + \frac{dy}{y} + \frac{dz}{z} = \;\;\; \Rightarrow\;\;\; xyz = c_1$
and
$\displaystyle x\, dx + y\, dy - dz = 0 \;\;\; \Rightarrow\;\;\; x^2 + y^2 -2 z = c_2$
giving the solution as
$\displaystyle x^2+y^2-2z = f(xyz)$
thanks for your solution,but one point ı am with trouble i did not understand your deduction which you did at here $\displaystyle \frac{dx}{x(y^2+z)} = \frac{dy}{-y(x^2+z)} = \frac{dz}{(x^2- y^2)z}$
to get this $\displaystyle \frac{dx}{x} + \frac{dy}{y} + \frac{dz}{z} = $
if you explain a little more ı will be gladfull to you.
Sure, let
$\displaystyle \frac{dx}{x(y^2+z)} = \frac{dy}{-y(x^2+z)} = \frac{dz}{(x^2- y^2)z}$
or
$\displaystyle \frac{dx}{dt} =x(y^2+z),\;\; \frac{dy}{dt} = -y(x^2+z),\;\; \frac{dz}{dt} = (x^2- y^2)z $
so
$\displaystyle \frac{1}{x}\,\frac{dx}{dt} =y^2+z,\;\; \frac{1}{y}\, \frac{dy}{dt} = -(x^2+z),\;\; \frac{1}{z}\, \frac{dz}{dt} = x^2- y^2 $
and if you add
$\displaystyle \frac{1}{x}\,\frac{dx}{dt} + \frac{1}{y}\,\frac{dy}{dt} + \frac{1}{x}\,\frac{dx}{dt} = 0$
or what I wrote earlier.
i discovered that u are solving wşth a different way from my teacher taught me,and i am at the start of pde hence i have some misunderstandable points left about your solution if you let me to ask you i will be happy,
how did you found c1 and c2 please explain me all steps ,than i will be capable of solving other questions with your way which is more understandable and easy to understand.thanks a lot for oyur helps danny.