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Just want a check to a question I've attempted. I have to find a general solution to this pde using the characteristic equation:

$ \frac{∂u}{∂x} - 4\frac{∂u}{∂y} - 3u = 0$

So I set $a=1, b=-4$ and $c=-3$ and got the characteristic coordinates:

$ξ = ((-4-\sqrt{19})x-y), η = (-4+\sqrt{19})x-y) $

And for the general solution I just put those two into $f_1(ξ) + f_2(η)$. Would appreciate any help on whether this is correct and if not, how to do it.

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Find a general solution of pde: $$\frac{∂u}{∂x} - 4\frac{∂u}{∂y} - 3u = 0\qquad(1)$$ We substitute $$u=ve^{3x}$$ For $v$ we get equation $$\frac{∂v}{∂x} - 4\frac{∂v}{∂y} = 0\qquad\quad(2)$$ Characteristic equation is $$\frac{dx}{1}=\frac{dy}{-4}$$ It solution is $$4x+y=C$$ Then solution of $(2)$ is $$v=f(4x+y)$$ and general solution of $(1)$ is $$u=f(4x+y)e^{3x}$$

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