Question. Given the following second order hyperbolic pde $$u_{tt} - a(x)u_{t} = u_{xx} + b(x) u_{x}+ c(x)u + d(x) \label{1}\tag{$\dagger$}$$ where $a(x)$, $b(x)$, $c(x)$ and $d(x)$ are smooth, is there a general method to find a solution $u(x,t)$ that is analytic in $t$ on a neighborhood of $t=0$, provided that the initial conditions $$ \begin{split} u(x,0)&=f(x) \\ \dfrac{\partial}{\partial t} u(x,0)&=g(x) \end{split} $$ are known?
Here we may write $$ u(x,t):=\sum_{n\ge 0}f_n(x)\frac{t^n}{n!} $$ Then the above pde yields a second order recurrence for the coefficients $f_n$. So with the initial conditions that give $f_0$ and $f_1$, all $f_n$ are determined and thus the solution $u(x,t)$ is unique. However, here I want an expression of $u(x,t)$ other than the above expressed by the recurrence for the coefficients $f_n$.
The PDE in question comes from my research in combinatorics. In many combinatorial problems, we often witness second order recurrences $f_{n+2}(x) = (*\cdots*)$ where $(*\cdots*)$ contains terms involving $f_{n+1}(x)$ and $f_n(x)$. Then, we may translate this recurrence into an equation of the ordinary or exponential genenrating function of $f_n(x)$, say $F(x,t)=\sum_{n\ge 0}f_n(x)t^n$ or $\sum_{n\ge 0}f_n(x)\frac{t^n}{n!}$. For example, if in the recurrence, we have a term $n^2 f_n(x)$, then in the resulting equation, we have a second derivative term $F_{tt}$ plus some first derivative terms; if in the recurrence, we have a term $n \frac{d}{dx}f_n(x)$, then in the resulting equation, we have a second derivative term $F_{xt}$, and so on. For many such problems, we may then make a change of variables to transform the resulting differential equation into the form \eqref{1}.
The above also explains why I do not want a solution expressed by determining the coefficients.
