How to solve following linear differential-difference equation? How to solve following linear differential-difference equation?
$$\frac{da_{n}(t)}{dt}=i k na_{n}(t)+G\left\{n(n-1)a_{n-2}(t)-a_{n+2}(t) \right\},~n=0,1,2,\ldots~~~~~(1)$$
where, k and G is a constant. And $i$ is the imaginary unit. The initial conditions are 
$$a_{n}(0)=\delta_{mn}=\begin{cases}
C_{0}~~(n=m)
\\
0~~(n\neq m)
\end{cases}$$ where, $C_{0}$ is a constant.
$a_{0}(t)$ and $a_{1}(t)$ is unknown function. I want to find $a_{n}(t)$. 
If $k=0$, then equation $(1)$ reduce to following equation:
$$
\frac{da_{n}(t)}{dt}=G\left\{n(n-1)a_{n-2}(t)-a_{n+2}(t) \right\},~n=0,1,2,\ldots.~~~~~(\mathrm{A})
$$
The general solution of equation $(\mathrm{A})$ is
$$
a_{n}(t)=C_{0}\frac{1}{\sqrt{\mathstrut 2\pi}}\left(\frac{n}{2} \right)!~\sqrt[]{\mathstrut 2^{n}}\left\{1+(-1)^{n+1} \right\}(\cosh{2Gt})^{-\frac{3}{2}}(\tanh{2Gt})^{-\frac{1}{2}(n-1)}.~~~~~(\mathrm{B})
$$
Equation $(\mathrm{B})$ satisfy following initial conditions:
$$
a_{n}(0)=\begin{cases}
C_{0}~~(n=1)
\\
0~~~~~(n\neq 1)
\end{cases}
~~~~~~~~~(\mathrm{C})
$$
I have tried Laplace transform to equaton $(1)$.
Multiplying both sides of equation $(1)$ by $\mathrm{e}^{-st}$ and then Integrating  for the interval $0$ to $\infty$  to obtain
$$
\int_{0}^{\infty}dt~\mathrm{e}^{-st}
\frac{da_{n}(t)}{dt}=i k n\int_{0}^{\infty}dt~\mathrm{e}^{-st}a_{n}(t)+G\left\{n(n-1)\int_{0}^{\infty}dt~\mathrm{e}^{-st}a_{n-2}(t)-\int_{0}^{\infty}dt~\mathrm{e}^{-st}a_{n+2}(t) \right\}.~~~~~(2)
$$
We define the $U_{n}(s)$
$$
U_{n}(s):=\int_{0}^{\infty}dt~\mathrm{e}^{-st}a_{n}(t).~~~~~(3)
$$
Then equation $(2)$ to be
$$
sU_{n}(s)-a_{n}(0)=iknU_{n}(s)+
G\left\{n(n-1)U_{n-2}(s)-U_{n+2}(s) \right\}.~~~~~(4)
$$
where, we use following integration by parts
$$
\int_{0}^{\infty}dt~\mathrm{e}^{-st}\frac{d a_{n}(t)}{dt}=sU_{n}(s)-a_{n}(0).
$$
Let's solve equation $(4)$ by using Z-transform.
First, we define unilateral Z-transform $W(s,z)$ as follows:
$$
\mathcal{Z}[U_{n}(s)]=
W(s,z):=\sum_{n=0}^{\infty} U_{n}(s)z^{-n}.~~~~~(5)
$$ 
Noting following relations
Differentiation & Time delay 
$$
\mathcal{Z}[n(n-1)U_{n-2}(s)]=
2z^{-2}W(s,z)-2z^{-1}\frac{\partial W(s,z)}{\partial z}+\frac{\partial^{2}W(s,z)}{\partial z^{2}},
~~~~~~~~(6)
$$
Time advance 
$$
\mathcal{Z}[U_{n+2}(s)]=z^{2}W(s,z)-z^{2}U_{0}(s)-zU_{1}(s),
~~~~~~~~~(7)
$$
Using propaty of $a_{n}(0)=\delta_{mn}$
$$
\mathcal{Z}[a_{n}(0)]=C_{0}z^{-m},
~~~~~~~~~~(8)
$$
we can transform equation $(4)$ as follows: 
$$
Gz^{2}\frac{\partial^{2}W(s,z)}{\partial z^{2}}+(-2G-ikz^{2})z\frac{\partial W(s,z)}{\partial z}
+\left\{z^{2}(-Gz^{2}-s)+2G \right\}W(s,z)+Gz^{3}\left\{zU_{0}(s)+U_{1}(s) \right\} +C_{0}z^{-m+2}=0.
~~~~~~(9)
$$
Equation $(9)$ is some kind of Bessel equation.
I'm trying to solve equation $(9)$.
 A: Hint
If:
$$
\mathcal{Z}[U_{n}(s)]=
W(s,z):=\sum_{n=0}^{\infty} U_{n}(s)z^{-n}
$$ 
Then:
$$
\mathcal{Z}[U_{n-2}(s)]=
\sum_{n=0}^{\infty} U_{n-2}(s)z^{-n}=z^{-2}\sum_{n=0}^{\infty} U_{n-2}(s)z^{-(n-2)}=z^{-2}\sum_{n=0}^{\infty} U_{n-2}(s)z^{-(n-2)}
$$ 
Now because $a_{n-1}=0$ and $a_{n-2}0$ (please check this assumption)
$$
\mathcal{Z}[U_{n-2}(s)]=
\sum_{n=0}^{\infty} U_{n-2}(s)z^{-n}=z^{-2}\sum_{n=0}^{\infty} U_{n-2}(s)z^{-(n-2)}=z^{-2}\mathcal{Z}[U_{n}(s)] = z^{-2}W(s,z) = \Gamma(s,z)
$$
And this is the time-shift property. I'll call the result $\Gamma$ because it will soon be convenient. Then:
$$
\gamma(s,n) =U_{n-2}(s) 
$$
Now look at
$$
\mathcal{Z}[n(n-1)U_{n-2}(s)]=\mathcal{Z}[n(n-1)\gamma(s,n)] =\mathcal{Z}[n^2\gamma(s,n)]- \mathcal{Z}[n\gamma(s,n)]
$$
From the differentiation property:
$$
\mathcal{Z}[n\gamma(s,n)] = -\frac{d}{dz} \left[ \Gamma(s,z)\right] =-\frac{\partial}{\partial z}\left[ \Gamma(s,z)\right] =-\frac{\partial}{\partial z}\left[ z^{-2}W(s,z)\right] =\\ 2z^{-3}W(s,z)-z^{-2}\frac{\partial W(s,z)}{\partial z}
$$
Now do it again to get  $\mathcal{Z}[n^2\gamma(s,n)] $. Then rebuild your equation 6.
