<s>[INST] How is $\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $ equivalent to $\frac{ y dx + xdy - zdz}{0}=\frac{ xdx - ydy -zdz}{0}$? In the book of PDE by Kumar, it is given that 

However, I couldn't figure out how is 
$$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 }  $$
is equivalent to 
$$\frac{ y dx + xdy - zdz}{0 } = \frac{ x dx - y dy -z dz}{ 0}  .$$
 [/INST] Some notes about solving PDE
$$z(x+y)z_x+z(x-y)z_y=x^2+y^2\qquad (1)$$


*

*see
https://www.math24.net/first-integrals-page-2/
Example 4

*In our case two independent first integrals is:
$$y^2+2xy-y^2=C_1, \quad z^2-2xy=C_2$$

*General solution of PDE $(1)$ is
$$z^2=2xy+f(y^2+2xy-x^2)$$

*Let $u=z^2$. Then from $(1)$ we get linear PDE
$$(x+y)u_x+(x-y)u_y=2x^2+2y^2\qquad (2)$$
with general solution
$$u=2xy+f(y^2+2xy-x^2)$$

*$u_p=2xy$ is particular solution of $(2)$

*$y^2+2xy-y^2=C_1$ is general solution of ODE
$$\frac{dx }{x+y} = \frac{dy}{x-y}.$$
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