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.

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    $\begingroup$ Try $A_n(x) = be^{rx}c^n$. You can get a relationship between the constants $c$ and $r$. $\endgroup$ – Michael Dec 22 '18 at 3:03

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 $(c_0,c_1)=(1,0)$, $(d_0,d_1) = (0,1)$.


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