How is $\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $ equivalent to $\frac{ y dx + xdy - zdz}{0}=\frac{ xdx - ydy -zdz}{0}$?

Question

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 $$\frac{ y dx + xdy - zdz}{0 } = \frac{ x dx - y dy -z dz}{ 0} .$$

Answer 1

$$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $$ Use the well known basic property of the fractions : $$\text{if}\quad \frac{A}{B}=\frac{C}{D} \quad \text{then}\quad \frac{A}{B}=\frac{C}{D}=\frac{c_1A+c_2C}{c_1B+c_2D}$$ $c_1,c_2$are arbitrary constants (not both nul}.

This property is valid for more fractions : $$\frac{A}{B}=\frac{C}{D}=\frac{E}{F}=\frac{c_1A+c_2C+c_3E}{c_1B+c_2D+c_3F}$$

In the case of Eq.$(1)$ with $c_1=y\quad;\quad c_2=x \quad;\quad c_3=-z$ : $$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } =\frac{ydx+xdy-zdz}{yz(x+y)+xz(x-y)-z(x^2+y^2)}=\frac{ydx+xdy-zdz}{0}$$ This implies $ydx+xdy-zdz=0$.

On the same way, with $c_1=x\quad;\quad c_2=-y \quad;\quad c_3=-z$ : $$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } =\frac{xdx-ydy-zdz}{xz(x+y)-yz(x-y)-z(x^2+y^2)}=\frac{xdx-ydy-zdz}{0}$$ This implies $xdx-ydy-zdz=0$.

Answer 2

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 $$\frac{dx }{x+y} = \frac{dy}{x-y}.$$