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I've just solved my first pde with the method of characteristics. The textbook I'm following gives many examples and this pde is one of them : $$x^2u_x+y^2u_y=(x+y)u$$

Applying the equation : $$\frac{dx}{a(x,y)}=\frac{dy}{b(x,y)}=\frac{du}{f(x,y)}$$

in this case we get: $$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{du}{(x+y)u}$$

From the last equation the author claims this:

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

How is this true?

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    $\begingroup$ It's called compendo-dividendo. $\endgroup$ – mattos Jun 23 '18 at 11:06
  • $\begingroup$ It's a very common identity $$ \frac{a}{b} + \frac{c}{d} = \frac{a+c}{b+d} $$ $\endgroup$ – Dylan Jun 23 '18 at 14:35
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It turned out to be very simple. We have :

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

Subtracting these two gives: $$dx-dy=\frac{(x^2-y^2)du}{(x+y)u}$$ $$=>\frac{dx-dy}{x^2-y^2}=\frac{du}{(x+y)u}$$

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