Solve the PDE: $u_{xx} - 3u_{xt} - 4u_{tt} = 0$ It is asked to solve the PDE
$$u_{xx} - 3u_{xt} - 4u_{tt} = 0$$
using a factorization, that consists in 
$$\left( \frac{\partial}{\partial x} - 4 \frac{\partial}{\partial t} \right) \left( \frac{\partial}{\partial x} + \frac{\partial}{\partial t} \right) u = 0$$
My attempt:
I called 
$$\left( \frac{\partial}{\partial x} + \frac{\partial}{\partial t} \right) = v$$ 
and solved 
$$ v_x - 4 v_t = 0$$
making the change of variables: $x' = x-4t, \; t' = -4x - t$.  This change of variables leave us with the following "ODE":
$$17 u_{x'} = 0 \Rightarrow u(x,t) = f(t') = f(-4x-t) = g(4x+t)$$
where g and f are arbitrary functions of one variable
For the second part, we have the PDE:
$$u_x+u_t = g(4x+t)$$
I made the following change of variables: $x' = x+t, t' = x-t$, and this leave us with the following "ODE":
$$2u_{x'} = g(4x+t)$$
The answer is 
$$ u(x,t) = c_1(4x+t)+c_2(x-t)$$
I understand that the last "ODE" give us the solution
$$u(x,t) = \int g(4x+t) dx' + c_2(x-t)$$
but why is the integral also a function of 4x+t? Where I commited a mistake?
Thanks in advance!
@Edit: Please, don't give me a solution using another technique that is not a factorization. I think my mistake, if there is one, consists on the variables that I am choosing.
 A: It will be helpful to view the "factorization" technique as separation of variables.
Assume $\,x = \eta+\xi ,\,$ and $\,t = \eta-4\xi  .\,$ 
Then
\begin{cases} 
x =  \eta + \xi\\ 
t = \eta-4\xi  
\end{cases}
\implies 
\begin{cases} 
\dfrac{\partial u}{\partial \xi}  = 
\dfrac{\partial u}{\partial x}\dfrac{\partial x}{\partial \xi} + 
\dfrac{\partial u}{\partial t}\dfrac{\partial t}{\partial \xi}  
= \dfrac{\partial u}{\partial x}  - 4 \dfrac{\partial u}{\partial t} 
\\ 
\dfrac{\partial u}{\partial \eta}  = 
\dfrac{\partial u}{\partial x}\dfrac{\partial x}{\partial \eta} + 
\dfrac{\partial u}{\partial t}\dfrac{\partial t}{\partial \eta}  
= \dfrac{\partial u}{\partial x}  + \dfrac{\partial u}{\partial t} 
\end{cases}
But then
\begin{multline}
\left( \frac{\partial}{\partial x} - 4 \frac{\partial}{\partial t} \right) 
\left( \frac{\partial}{\partial x} + \frac{\partial}{\partial t} \right) u 
=
\left( \frac{\partial}{\partial \xi}   \right) 
\left( \frac{\partial}{\partial \eta} \right) u 
= u_{\xi\eta}  = 0
\end{multline}
We can write 
\begin{alignat}{1}
u_{\xi\eta}  = 0 & \implies u_{\xi} & = \int u_{\xi\eta} \,d\eta = c_1\left(\xi\right), \quad
u_{\eta} = \int u_{\xi\eta} \,d\xi = c_2\left(\eta\right),&
\\
&\implies   u & = \iint u_{\xi\eta} \,d\eta \,d\xi = \int c_1\left(\xi\right) \, d\xi 
= C_1\left(\xi\right) + C_2\left(\eta\right)&
\\
&\implies u_\eta  & = \dfrac{d C_2}{d\eta}  = \int u_{\xi\eta} \,d\eta = c_2\left(\eta\right)
&\\ 
&\implies u & = C_1\left(\xi\right) +  C_2\left(\eta\right)&
\end{alignat}
In that case it is clear that 
$$
u = C_1\left(\xi\right) +  C_2\left(\eta\right) = 
C_1\big(x-t \big) + C_2\big(4x+t \big) 
$$
Since $\,c_1,\,$ $\,c_2,\,$ and $\,c_3\,$ are arbitrary functions, we can rewrite the last equation as 
$$
\bbox[5pt, border:2.5pt solid #FF0000]{
u =  C_1 \left(x-t \right) + C_2 \left(4x+t \right) 
}
$$
