So we have the following problem
$$ U_{xx} - 3U_{xt}-4U_{tt}=0 \tag{1}\\ U(x,0)=x^2 \\ U_{t}=e^x $$
So Solving this by factorising and using a change of variables we get
$$U_{xx} - 3U_{xt}-4U_{tt}=\left(\frac{\partial}{\partial x}-4\frac{\partial}{\partial t}\right)\left( \frac{\partial}{\partial x}+\frac{\partial}{\partial t}\right)U=0$$ Define $$ W(x,t) := \left( \frac{\partial}{\partial x}+\frac{\partial}{\partial t}\right)U $$ we then have two consecutive first order pde's witch constant coefficients which can then be solved using the method of characteristics to give. $$\begin{equation}U(x,t)=(x^2+\frac{1}{4}t^2)+\frac{4}{5}e^{x}(e^{\frac{t}{4}}-e^{-t}) \tag{2}\end{equation} $$
So all the above I did, but I decided to try and solve $(1)$ using D'Alembert's solution, namely;
$$U(x,t) = \frac{1}{2}\big(f(x+ct)+f(x-ct) + \frac{1}{2c}\int^{x+ct}_{x-ct}g(s)ds \tag{3}$$ Where $$ \ \ U(x,0)=x^2=f(x), \\ U_{t}=e^x=g(x)$$
So I started by using the following change of vairables $$x'=x, \tag{4} \\t'=-\frac{b}{a}x+t$$ so that $$\frac{\partial}{\partial t'}=\frac{\partial}{\partial t}, \\\frac{\partial}{\partial x'} = \frac{\partial}{\partial x} + \frac{b}{a}\frac{\partial}{\partial t}$$
hence with $a=1, \ b=-\frac{3}{2}\ and \ c=-4,$ $(1)$ becomes, $$aU_{xx}+2bU_{xt}+cU_{tt}=U_{xx} - 3U_{xt}-4U_{tt}= U_{x'x'} -\frac{25}{4}U_{t't'}$$
it is now in the form of the wave equation in $1+1$ dimensions with $c^2=\frac{25}{4}$ so we can use D'Alembert Solution $(3)$. From here I converted the coordinates for the initial conditions $$U(x,0) = U(x',0)=x'^2$$ since $\frac{\partial}{\partial t'}=\frac{\partial}{\partial t}$ we also get $$U_{t}=U_{t'}=e^{x'}$$ Therefore in terms of $x'$ and $t'$, $(3)$ becomes $$U(x',t')=\frac{1}{2}\big((x'+ct')^2 +(x'-ct')^2\big) + \frac{1}{2c}\big( e^{x'+ct'}- e^{x'-ct'}\big)$$
Plugging $(4)$ back in to get the solution in terms of the original variables we get $$U(x,t) = \frac{1}{2}\big(2x^2 +2c^2(\frac{3}{2}x+t)^2\big) +\frac{1}{2c}\big( e^{x+c(\frac{3}{2}x+t)} - e^{x-c(\frac{3}{2}x+t)}\big) \tag{5}$$
but $(5) \neq (2)$ and I dont know where I went wrong.
So basically that is my question, where did I go wrong?.