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