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.

• Try $A_n(x) = be^{rx}c^n$. You can get a relationship between the constants $c$ and $r$. – 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 $$(c_0,c_1)=(1,0)$$, $$(d_0,d_1) = (0,1)$$.