Solve $u_{xx}-3u_{xt}-4u_{tt}=0$ where $u(x,0)=x^2$ and $u_t(x,0)=e^x$ Solve
$$u_{xx}-3u_{xt}-4u_{tt}=0$$
where $u(x,0)=x^2$ and $u_t(x,0)=e^x$.

My workings so far: I have factored the differential equation in the following way:
$$(\delta_x-4\delta_t)(\delta_x+\delta_t)=0$$
where $\delta_x=\frac{\delta}{\delta x}$ etc. Now if we let $v$ be the solution to $(\delta_x+\delta_t)u$ then we have the following two equations:
\begin{eqnarray*}
(\delta_x+\delta_t)u=u_x+u_t=v\\
(\delta_x-4\delta_t)v=v_x-4v_t=0
\end{eqnarray*}
Now for $v$ we simply find
$$v=h(t+4x)$$
where $h$ is an arbitrary function of one variable. Now what remains is find $u$ such that
$$u_x+u_t=h(t+4x)$$
I am stuck here, I thought about making a change of variables $\zeta=x+t$ and $\eta=x-t$ and thus using the product rule to show that $u_x=u_{\zeta}+u_{\eta}$ and $u_t=u_{\zeta}-u_{\eta}$ and thus $u_x+u_t=2u_{\zeta}$ and we need to solve
$$u_{\zeta}=h(t+4x)$$
(I left out the factor 2 because $h$ is an arbitrary function). Do I simply integrate now and conclude
$$u=f(\zeta)h(t+4x)+g(\eta)=f(x+t)h(t+4x)+g(x-t)$$
This seems wrong to me...

Some help would be greatly appreciated!
Note that, since the pde is with constant coefficients, then we proceed as 

$$u_{xx}-3u_{xt}-4u_{tt}=0\implies(\delta_x-4\delta_t)(\delta_x+\delta_t)u=0$$

$$\implies(\delta_xu-4\delta_tu=0)\cup(\delta_xu+\delta_tu=0)$$

$$\implies\left\{u(x,t)=F(4x+t)\right\}\cup\left\{u(x,t)=G(t-x)\right\}$$

$$\implies u(x,t)=F(4x+t)+G(t-x)$$

$$\implies u(x,t)=f\left(x+\dfrac{t}{4}\right)+g(x-t)$$

$$u(x,0)=x^2\implies f(x)+g(x)=x^2$$

$$u_t(x,0)=e^x\implies\dfrac{f'(x)}{4}-g'(x)=e^x\implies\dfrac{f(x)}{4}-g(x)=e^x+c$$

$$\therefore f(x)=\dfrac{4x^2+4e^x+4c}{5},g(x)=\dfrac{x^2-4e^x-4c}{5}$$

$$\therefore u(x,t)=\dfrac{4\left(x+\dfrac{t}{4}\right)^2+4e^{x+\frac{t}{4}}+(x-t)^2-4e^{x-t}}{5}$$
I solve with free CAS Maxima. We use Laplace transform method.



Answer:

$$u=\frac{4}{5}e^{x+\frac{t}{4}}-\frac{4}{5}e^{x-t}+x^2+\frac{t^2}{4}$$