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

Note that, since the pde is with constant coefficients, then we proceed as

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

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

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

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

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

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

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

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

$$\therefore u(x,t)=\dfrac{4\left(x+\dfrac{t}{4}\right)^2+4e^{x+\frac{t}{4}}+(x-t)^2-4e^{x-t}}{5}$$