PDE : $x^2 z_x + y^2 z_y = z(x+y)$

Question

Solving the PDE $$x^2 z_x + y^2 z_y = z(x+y)$$

I came across an error in my calculations which I cannot find :

$$\frac{\mathrm{d}x}{x^2}=\frac{\mathrm{d}y}{y^2}=\frac{\mathrm{d}z}{z(x+y)}$$

The first integral curve is given as :

$$\frac{\mathrm{d}x}{x^2}=\frac{\mathrm{d}y}{y^2}\implies z_1 = \frac{1}{x} - \frac{1}{y}$$

Now, for the second integral curve, I use the following differential subtraction trick to get rid of the $(x+y)$ term :

$$\frac{\mathrm{d}x-\mathrm{d}y}{x^2-y^2}=\frac{\mathrm{d}z}{z(x+y)}$$ $$\Rightarrow$$ $$\frac{\mathrm{d}(x-y)}{x-y} = \frac{\mathrm{d}z}{z}$$ $$\implies$$ $$z_2 = \frac{z}{x-y}$$

And thus the general solution is : $$z_2=F(z_1)\Rightarrow z(x,y)=(x-y)F\bigg(\frac{1}{x}-\frac{1}{y}\bigg)$$ where $F$ is a $C^1$ function of $x$ and $y$.

Wolfram Alpha though, states that the solution is $z(x,y)=xyF(1/x-1/y$), which means that I have found the integral curve $z_2$ wrong. I cannot find any fault in my calculations though. The correct integral curve should be : $$z_2 = \frac{z}{xy}$$ How would one come to this calculation though ?

I would really appreciate your help

Answer 1

From

$$ \frac{1}{x}-\frac{1}{y}=C_1 \Rightarrow y-x=C_1 x y $$

From

$$ \frac{dy-dx}{y-x}=\frac{dz}{z}\Rightarrow y-x = C_2 z $$

hence

$$ C_2 z = C_1 x y \Rightarrow C_4 = \frac{z}{x y} $$

Answer 2

Some algebra helps too:

$(x-y)F\bigg(\dfrac{1}{x}-\dfrac{1}{y}\bigg)=xy\left(\dfrac{x-y}{xy}\right)F\bigg(\dfrac{1}{x}-\dfrac{1}{y}\bigg)=xy\left(\dfrac{1}{y}-\dfrac{1}{x}\right)F\bigg(\dfrac{1}{x}-\dfrac{1}{y}\bigg)=xyG\left(\dfrac{1}{x}-\dfrac{1}{y}\right)$

with $G\left(\dfrac{1}{x}-\dfrac{1}{y}\right)=-\left(\dfrac{1}{x}-\dfrac{1}{y}\right)F\bigg(\dfrac{1}{x}-\dfrac{1}{y}\bigg)$

So, they are essentialy the same.

Answer 3

$$dt =\frac{\mathrm{d}x}{x^2}=\frac{\mathrm{d}y}{y^2}=\frac{\mathrm{d}z}{z(x+y)}$$ is equivalent to $$ dt =\frac{zy dx + zx dy - xy dz }{0 },$$

hence the numerator is constant, which only depends on $C_1 = \frac{ 1}{x } - \frac{ 1}{ y}. $

This implies by integrating each side $$F(C_1) = zyx + zxy - xyz = zxy,$$ Thus, $$z = \frac{F(\frac{ 1}{x } - \frac{ 1}{ y}) }{ xy} .$$