# Find general solution to PDE using characteristic equation

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.

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 $$u=f(4x+y)e^{3x}$$