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The PDE is the following :

$$(xy)u_x+(2xy)u_y=(x+y)u$$

To be honest, I haven't quite understood the method and it seems to be used in different ways and that confuses me(in some solutions I see parametrizations going on). I managed to apply it in some problems using the following characteristics equations, which in this problem are written like this :

$$\frac{dx}{xy}=\frac{dy}{2xy}=\frac{du}{(x+y)u}$$

From the first equality we get:

$$\frac{dy}{dx}=2$$ $$=>y=2x+C_1=>C_1=y-2x$$

We can also get this:

$$\frac{du}{dx}=\frac{(x+y)u}{xy}=(1/y+1/x)u$$ Integrating I get something strange (from what I've seen at least).

$$ln|u|=x/y+ln|x|+C_2$$

Am on the right track? Can I use my initial equations to get a better relationship for C2? By cancelling (x+y) perhaps?

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  • $\begingroup$ You're fine. The last step is to just take the exponential of both sides to get $$ u = Cxe^{x/y} = xe^{x/y}f(y-2x) $$ $\endgroup$ – Dylan Jun 23 '18 at 14:33
  • $\begingroup$ I did think of that but how does the absolute value disappear? I had the same problem in other examples but I found another way around which didn't require producing an absolute value. I couldn't avoid it on this one. $\endgroup$ – John Katsantas Jun 23 '18 at 14:40
  • $\begingroup$ @Dylan Maybe I've made a mistake, as far as I can tell your solution doesn't satisfy the PDE. The integration at $$u' = \left( \frac{1}{x} + \frac{1}{y} \right)u$$ is incorrect as $y = y(x)$. $\endgroup$ – Mattos Jun 23 '18 at 14:58
  • $\begingroup$ Ok, I made a mistake. The integration should have $dy=2dx$. But then @Mattos your solution is also not correct. It should be $$ u = x\sqrt{y}f(y-2x) $$ $\endgroup$ – Dylan Jun 23 '18 at 20:15
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    $\begingroup$ @JohnKatsantas I wouldn't worry too much about the absolute value. $|x|$ produces either a positive or negative sign which gets absorbed into the arbitrary function $\endgroup$ – Dylan Jun 23 '18 at 20:17
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I should clarify that I'm also not an expert in the method, but I can tell you what went wrong.

Your first part up to $c_1 = y-2x$ is correct

For the second integral, you have $dy=2dx$, therefore

$$ \frac{du}{u} = \left(\frac{1}{y} + \frac{1}{x}\right)dx = \frac{1}{2y}dy + \frac{1}{x}dx $$

Integrating the above

$$ \ln |u| = \frac12 \ln |y| + \ln|x| + c_2 $$

Taking the exponential

$$ |u| = e^{c_2}|x|\sqrt{|y|} \implies u = \pm e^{c_2}x\sqrt{|y|} = Cx\sqrt{|y|} $$

The last step is to take $C = f(c_1)$ so that

$$ u = x\sqrt{y} f(y-2x) $$

You can usually drop the absolute value on $|y|$ for convenience, unless a boundary condition is given.

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