I'm supposed to solve $u_{xx}-3u_{xt}-4u_{tt}=0$ with initial conditions $u(x,0)=x^2$ and $u_t(x,0)=e^x$.
So I factored the problem into $(u_x-4u_t)(u_x + u_t)$ and set each equal to 0 and found the 2 solutions to be $(x,t)=f(x+t/4)$ and $u(x,t)=g(x-t)$
Then I get $u(x,t)=1/2[\phi(x+t/4) + \phi(x-4)]$ + 1/2c $\int_{x-t}^{x+t/4} e^{s} ds$
=1/2[$(x+t/4)^2 + (x-t)^2$] + 1/2c[$e^{x+t/4}-e^{x-t}$]
I multiply out the first part then and get a solution of:
1/2[$5/2xt-15/16t^2$]+1/2c[$e^{x+t/4}-e^{x-t}$]
But the book has a solution of $x^2 + t^2/4 + 4/5[e^{x+t/4}-e^{x-t}]$
I don't understand how they get 4/5 in front of the exponentials and how $1/2[\phi(x+t/4) + \phi(x-4)]$ turns into $x^2 + t^2/4$ . Can anyone see where I went wrong?
