I am aware that the PDE is already solved in this thread.
But, I want to know how that person got from $u_x+u_t=g(4x+t)$ to $2u_{x'}=g(4x+t)$ by using the change of variables $x′=x+t,t′=x−t$. I don't understand the $2u_{x'}$ part. Finally, after doing all that, how do you arrive at the final solution?
I prefer not following external links unless I have to, so I hope the following might help.
Since $a^2-3 ab-4b^2=(a+b)(a-4b)$ we can (not rigorously) write: $$ \left( \frac{\partial^2}{\partial x^2} -3 \frac{\partial^2}{\partial x \partial t}-4 \frac{\partial^2}{\partial t^2}\right)u=\left( \frac{\partial}{\partial x} +\frac{\partial}{\partial t}\right)\left( \frac{\partial}{\partial x} -4\frac{\partial}{\partial t}\right)u=0$$
And obtain two 1st order PDEs:
$$\left( \frac{\partial}{\partial x} +\frac{\partial}{\partial t}\right)u_1=0$$
$$\left( \frac{\partial}{\partial x} -4\frac{\partial}{\partial t}\right)u_2=0$$
Where we are searching for the solution as:
$$u(x,t)=A u_1(x,t)+B u_2(x,t)$$
By the form of the above equations, it's clear that we should search for solutions as:
$$u_1(x,t)=f(x-t)$$
$$u_2(x,t)=g(4x+t)$$
You can check all of this by direct substitution in the original PDE.