Solve the PDE: $u_{xx} - 3u_{xt} - 4u_{tt} = 0$ [duplicate]

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

$$ \bbox[5pt, border:2.5pt solid #FF0000]{ u = C_1 \left(x-t \right) + C_2 \left(4x+t \right) } $$