I have a partial differential equation :
$u_{xx}-3u_{xt}-4u_{tt}=0$ with initial and boundary conditions: $u(x,0)=x^2$, $u_t(x,0)=e^x$. We need to find $u(x,t)$. The hint is : Factor the operator as was done with the wave equation $u_{tt}= c^2u_{xx}$.
Here is my thought process. Operator factoring is exactly like factoring from elementary school algebra. So I can pretend for a moment that
$u_{xx}-3u_{xt}-4u_{tt}=>(x^2-3xt-4t^2)=(x-4)(x+1) $
So
$u_{xx}-3u_{xt}-4u_{tt}= (\partial_x -4\partial_t)(\partial_x+\partial_t)u=0$
So the last step above is using the suggested hint of operator factoring.
From the general solution of wave equations we know it will have the form:
$u(x,t)=f(something1) + g(something2)$
Using the factored term: $(\partial_x -4\partial_t)$ we know that $something1 = x-4t$. Using the factored term $(\partial_x+\partial_t)$ we know that $something2 = x + t$.
Therefore the solution is:
$u(x,t)=f(x-4t) + g(x+t)$
Then I use the initial and boundary conditions giving me:
$u(x,0)=f(x)+g(x)=x^2$ $u_t(x,0)=g'(x)-4f'(x)=e^x$
However, I am unable to get the correct answer from here:
$\frac{4}{5} (e^{x+t/4}-e^{x-t}) + x^2+\frac{1}{4}t^2$