Solution to a particular Wave Equation Consider the partial differential equation
\begin{align}
\frac{1}{c^{2}} \, \frac{ \partial^{2} U}{\partial t^{2}} &= \frac{\partial^{2} U}{\partial x^{2}} + x \, \frac{\partial U}{\partial x} + \alpha \\
U(0,t) &= p \\
U(a,t) &= q \\
U(x,0) &= h(x) \\
\end{align}
It can be determined that the time component has a solution of the form 
\begin{align}
A_{1} \cos(\lambda c t) + B_{1} \sin(\lambda c t).
\end{align}
The question here is how to find the eigenvalues $\lambda$ and the solution to the $x$ component of the equation. What is the most appropriate method to use?

This is by no means a complete answer, but here are some notes on my try to solve the problem which might be useful for you or other people that wants to try to solve it.

We start by scaling the time coordinate $ct \to t$ to get
$$U_{tt} = U_{xx} + x U_x + \alpha$$
Since it was not specified we will for simplicity try to solve the PDE on the domain $(t,x)\in[0,\pi]\times [-\infty,\infty]$.
First we start by getting ridd of $\alpha$ by taking $U = \frac{\alpha t^2}{2} + \tilde{U}$. Next expand $\tilde{U}$ in a Fourier series $\tilde{U}(x,t) = \sum_{k=-\infty}^\infty U_k(x) e^{ikt}$ on the interval $[-\pi,\pi]$ to get the equations
$$U_k'' + x U_k'  + k^2 U_k = 0$$
With $U_k = Z_k e^{-\frac{x^2}{4}}$ we get
$$Z_k'' + \left(k^2-1 + \frac{1}{2}-\frac{x^2}{4}\right)Z_k = 0$$
The solutions for $Z$ are known as the Parabolic Cylinder functions: $D_{k^2-1}(x)$ and $D_{-k^2}(ix)$. $D_n$ is a terrible function to work with unless $n$ is a non-negative integer for which
$$D_n(x) \propto e^{-\frac{x^2}{4}}H_n\left(\frac{x}{\sqrt{2}}\right)$$
where $H_n$ are the well known Hermite polynomials. For negative $n$, $D_n$ is related to the error function. The full solution can formally be written
$$U(x,t) = \frac{t^2}{2}\alpha + \sum_{k=-\infty}^\infty \left[c_k D_{k^2-1}(x)  + d_kD_{-k^2}(ix)\right]e^{ikt-\frac{x^2}{4}}$$
for some (complex) numbers $c_k,d_k$. Before applying boundary condition it is useful to expand $t^2$ in a Fourier series on $[-\pi,\pi]$: 
$$t^2 = \frac{\pi^2}{3} + \sum_{k=-\infty,k\not=0}^\infty \frac{2(-1)^k}{k^2}e^{ikt}$$
Using this the boundary conditions can be written
$$p - \frac{\alpha\pi^2}{6} = \sum_{k=-\infty,k\not=0}^\infty \frac{\alpha(-1)^k}{k^2}e^{ikt} + \sum_{k=-\infty}^\infty \left[c_k D_{k^2-1}(0)  + d_kD_{-k^2}(0)\right]e^{ikt}$$
$$q - \frac{\alpha\pi^2}{6} = \sum_{k=-\infty,k\not=0}^\infty \frac{\alpha(-1)^k}{k^2}e^{ikt} + \sum_{k=-\infty}^\infty \left[c_k D_{k^2-1}(a)  + d_kD_{-k^2}(ia)\right]e^{ikt-\frac{a^2}{4}}$$
so by the uniqueness of Fourier series we must have
$$p-\frac{\alpha\pi^2}{6} = c_0 D_{-1}(0)  + d_0D_{0}(0)$$
$$q-\frac{\alpha\pi^2}{6} = \left[c_0 D_{-1}(a)  + d_0D_{0}(ia)\right]e^{-\frac{a^2}{4}}$$
and
$$c_k D_{k^2-1}(0)  + d_kD_{-k^2}(0) = \left[c_k D_{k^2-1}(a)  + d_kD_{-k^2}(ia)\right]e^{-\frac{a^2}{4}} = -\frac{\alpha(-1)^k}{k^2}$$
for $k\not= 0$.
Finally, applying the initial condition $U(x,0) = h(x)$ and using the orthogonality relation $\int_{-\infty}^\infty D_n(x)D_m(x)\,{\rm d}x = \delta_{nm}n!\sqrt{2\pi}$ we get
$$c_{n}  = \frac{1}{n!\sqrt{2\pi}}\int_{-\infty}^\infty{\rm d}x\left\{ h(x)e^{\frac{x^2}{4}}D_{n^2-1}(x) - \sum_{k=-\infty}^\infty \left[d_kD_{-k^2}(ix) D_{n^2-1}(x)\right]\right\}$$
This provides a closed system for $c_k,d_k$ which I have not been able to solve yet. It looks horrible, but perhaps there is some nice properties of the cylinder functions that can be applied to simplify it.
Another issue if whether the last integral in the $c_n$ expression converges. The reason I picked $[-\infty,\infty]$ as the domain for $x$ was the orthogonality condition so it might be better to instead consider a finite domain like $[0,a]$, but then I don't know of any orthogonality conditions to use.
