$u_{xx} − 7u_{tx} + 12u_{tt} = 0$ (Partial Differential Equations) $u_{xx} − 7u_{tx} + 12u_{tt} = 0$, $−1<x<1$, $t>0$, $u(0,x) = x^2$, $u_t(0,x) = e^x$ for x≥0
Attempt (Factorization Method): Factoring the differential operator,
$0 = (∂^2/∂x^2)u - 7(∂/∂x)(∂/∂t)u + 12(∂^2/∂t^2)u
= (∂^2/∂x^2 + 2(∂/∂x)(∂/∂t) + ∂^2/∂t^2)(u – (7/2)u + 12u)
= (∂/∂x + ∂/∂t)^2(19/2)u
= ∇^2(19/2)u
= ∇^2U(x,t)$ where $U = (19/2)u$
$∂^2U/∂x^2 + ∂^2U/∂t^2 = 0$
$U(x,t) = f(x)g(t)$ where $f(x) = U(0,x) = (19/2)u(0,x) = 19x^2/2$
$(∂^2/∂x^2)(f(x)g(t)) + (∂^2/∂t^2)(f(x)g(t)) = 0$
$f''(x)g(t) + f(x)g''(t) = 0$
Set $g(t) = a(x)sin(t) + b(x)cos(t)$,
$g''(t) = -a(x)sin(t) – b(x)cos(t)$
$f''(x)(a(x)sin(t) + b(x)cos(t)) - f(x)(a(x)sin(t) + b(x)cos(t)) = 0$
$(a(x)sin(t) + b(x)cos(t))(f''(x) – f(x)) = 0$
$(a(x)sin(t) + b(x)cos(t))(19 – 19x^2/2) = 0$
$(a(x)sin(t) + b(x)cos(t))(2 – x^2) = 0$
I heard that to solve the Laplace equation $∇^2U(x,t) = 0$ we needed to rearrange the terms to obtain $f''(x)/f(x) = -g''(t)/g(t)$ but how does this help us?
But then we get a(x) = b(x) = 0 so that g(t) = 0 and thus U = u = 0 which doesn't satisfy the initial condition.
The professor said to use the coordinate method; where does that occur?
 A: I don't think that is a factorisation: you start with a second order equation and end with an another with the same order. If you need to follow the idea of factorizing the equation, this is:
The equation factorizes: $(\partial_x-4\partial_t)(\partial_x-3\partial_t)u=0$. Separable in two equations of first order.
So, we have that a solution is a sum of two single argument twice differentiable functions, $f$ satisfying the first factor and $g$ satisfying the second: $(\partial_x-4\partial_t)f=0$ and $(\partial_x-3\partial_t)g=0$
The general solution is: $u(t,x)=f(4x+t)+g(3x+t)$
From here we impose the boundary conditions. Some formal manipulation will be enough.

Added
$u(0,x)=x^2=f(4x)+g(3x)\tag 1$
$u_t(0,x)=e^x=f'(4x)+g'(3x)\tag 2$
Differentiating $(1)$ $4f'(4x)+3g'(3x)=2x\implies g'(3x)=(2/3)x-(4/3)f'(4x)$
Substituting in $(2)$ $f'(4x)+(2/3)x-(4/3)f'(4x)=e^x\implies f'(4x)=2x-e^x$
$$f'(x)=x/2-3e^{x/4}$$
Integrating, $f(x)=x^2/4-12e^{x/4}+C$ and again with $(1)$
$g(3x)=x^2-4x^2+12e^x$ thus, $g(x)=-x^2/3+12e^{x/3}$
But by $(1)$, $C=0$
Finally,
$$u(t,x)=(4x+t)^2/4-(3x+t)^2/3+12(e^{(3x+t)/3}-e^{(4x+t)/4})$$
A: Even easier. Since the coefficients of your PDE are constant, let $u = C \exp(\alpha x + \beta t)$ for some reals $\alpha$ and $\beta$. Plugging this back into your equation leads to
$$ (\alpha/\beta)^2 - 7 (\alpha/\beta) + 12  = 0$$
which has as solution $\alpha/\beta = 4$ or $\alpha/\beta = 3$. Thus, you can express your solution as 
$$ u(x,t) = C_1 \exp[\beta ( 4 x  + t)] + C_2 \exp[\beta(3x + t) ]$$
with $C_i$ constants of integration. This solution is valid over all possible ranges of $\beta$. Moreover, the superposition of all possible solutions for every $\beta$ is also a solution so you can integrate over $\beta$ and say:
$$u(x,t) = C_1 \int_\beta \exp[\beta ( 4 x  + t)] \mathrm{d} \beta + C_2 \int_\beta \exp[\beta ( 3 x  + t)] \mathrm{d} \beta = F(4x+t) + G(3x+t) $$ where $F$ and  $G$ are arbitrary functions of their arguments. You can see the integrals as some Fourier-like transforms of functions of the arguments $4x+t$ and $3x+t$.

I know the last sentence is not very rigorous, but I'm afraid I can't do better on explaining it. Sorry about that.
