How to solve linear differential-difference equation? Given a linear differential-difference equation:
$$A_{n+2}+\partial A_{n+1}+\partial^2 A_n=0,$$
where $A$ is a function of $n$ and $x$, and $\partial$ represents the derivative about $x$.
How to solve this equation? The general case is this form
$$A_{n+2}+P_1 A_{n+1}+P_2 A_n=0,$$
where $P_1,P_2$ are differential operator depending on function of $x$.
I have tried to set $A_n=B^nA_0$, where $B$ is a pseudo differential operator and $A_0$ is a function of $x$, then I obtain
$$B^2+P_1B+P_2=0.$$
After solving $B$, we have the solution of $A_n$. I don't know whether this is correct and how to do next. 

Given any infinitely differentiable functions $A_0(x), A_1(x)$, we can proceed to find $A_n(x)$ for all $n \in \{0, 1, 2, ...\}$. 
Given $A_0(x)=f(x), A_1(x)=g(x)$, where $f, g$ are given infinitely differentiable functions,  a general solution is 
$$ A_n(x) = c_nf^{(n)}(x) + d_ng^{(n-1)}(x) \quad \forall n \in \{0, 1, 2, ...\} $$ 
where $f^{(n)}(x)$, $g^{(n)}(x)$ represent the $n$th derivative (and we use the convention $f^{(0)}= f$, $g^{(0)}=g$, $g^{(-1)}= 0$), and $c_n, d_n$ are solutions to the linear recurrence relation
\begin{align*}
c_{n+2} &=-c_{n+1} - c_n \quad \forall n \in \{0, 1, 2, ...\} \\
d_{n+2} &=-d_{n+1} - d_n \quad \forall n \in \{0, 1, 2, ...\} 
\end{align*}
with initial conditions $(c_0,c_1)=(1,0)$, $(d_0,d_1) = (0,1)$.
