$\frac{∂u}{∂x} − 4\frac{∂u}{∂y} − 3u = 0.$
I've tried $u=f(x+\alpha)$ to seek a solution of the form $u=g(x)f(x+4y)$
Which gives $\frac{du}{dx}=g'f+f'g$ and $\frac{du}{dy}=4gf'$.
Subbing this back into the original equation gives $g'f+5gf'-3gf=0$
However I think this is wrong as I have not yet learnt how to solve an equation of this form. Any help is appreciated.
This equation is easily solved with the method of characteristics.
In general, for $a\frac{\partial u}{\partial x}+b\frac{\partial u}{\partial y}=c$, is possible find the characteristic curves by solving the system of ODE's $\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}$ and then spreading those curves to the surface solution. So, we have to solve this system of coupled ODE's:
$$\frac{dx}{1}=-\frac{dy}{4}=\frac{du}{3u}$$
Taking the first two ratios: $4dx=-dy$, that leads to the first relation $4x=-y+c_1$ or $4x-y=c_1$
Taking the first and third, $3dx=\frac{du}{u}$ thus, $3x=\ln u + c'_2$ or $c_2e^{3x}=u$
$c_1$ and $c_2$ have to be somehow related, so is: $c_2=f(c_1)$ being f an arbitrary single variable prefereably continuous function, a function "driving" the characteristics to form a surface solution. $c_2=f(4x+y)$ And finally:
$$u=f(4x+y)e^{3x}$$
Trying to have a sensible habit, I've checked it and satisfies the equation.
Anyway, following the tip we can solve it too. Let us try $u(x,y)=g(x)f(4x+y)$ We have:
$$g'f+4gf'-4gf'-3gf=0\;or\;(g'-3g)f=0$$
Then or $f=0$ or $g'-3g=0$. Solving it we have $g(x)=c_3e^{3x}$ ($c_3$ can be eventually assimilated to (the arbitrary) f).