Solution to a second order recurrence relation with non constant coefficient I have the following equation:
$(aq^n+b+s)C_n(s)=aq^nC_{n+1}(s)+bC_{n-1}(s)$ , for $n\geq1$.
$C_0(s)=1$ ,  and for all $s\geq 0$ we have $0\leq C_n(s)\leq1$.
$a>0$,   $b>0$, $0\leq q\leq1$.
In fact, $C_n(s)$ is a Laplace Transform of a non-negative RV, for all n.
Any thoughts on how to solve this equation?
Thanks!
 A: Case $1$: $q=0$
Then $(b+s)C_n(s)=bC_{n-1}(s)$
$C_n(s)=\dfrac{bC_{n-1}(s)}{b+s}$
$C_n(s)=\dfrac{b^n\Theta(n)}{(b+s)^n}$ , where $\Theta(n)$ is an arbitrary periodic functions with unit period
$C_0(s)=1$ :
$\Theta(0)=1$
$\therefore C_n(s)=\dfrac{b^n\Theta(n)}{(b+s)^n}$ , where $\Theta(n)$ is an arbitrary periodic functions with unit period and $\Theta(0)=1$
Case $2$: $q=1$ 
Then $(a+b+s)C_n(s)=aC_{n+1}(s)+bC_{n-1}(s)$
$aC_{n+1}(s)-(a+b+s)C_n(s)+bC_{n-1}(s)=0$
The auxiliary equation is
$a\lambda^2-(a+b+s)\lambda+bC_{n-1}(s)=0$
$\lambda=\dfrac{a+b+s\pm\sqrt{(a+b+s)^2-4ab}}{2a}$
$\therefore C_n(s)=\begin{cases}\Theta_1(n)\left(\dfrac{a+b+s+\sqrt{(a+b+s)^2-4ab}}{2a}\right)^n+\Theta_2(n)\left(\dfrac{a+b+s-\sqrt{(a+b+s)^2-4ab}}{2a}\right)^n&\text{when}~(a+b+s)^2-4ab\neq0\\\Theta_1(n)n\left(\dfrac{a+b+s}{2a}\right)^n+\Theta_2(n)\left(\dfrac{a+b+s}{2a}\right)^n&\text{when}~(a+b+s)^2-4ab=0\end{cases}$ , where $\Theta_1(n)$ and $\Theta_2(n)$ are arbitrary periodic functions with unit period
$C_0(s)=1$ :
$\begin{cases}\Theta_1(0)+\Theta_2(0)=1&\text{when}~(a+b+s)^2-4ab\neq0\\\Theta_2(0)=1&\text{when}~(a+b+s)^2-4ab=0\end{cases}$
$\therefore C_n(s)=\begin{cases}\Theta_1(n)\left(\dfrac{a+b+s+\sqrt{(a+b+s)^2-4ab}}{2a}\right)^n+\Theta_2(n)\left(\dfrac{a+b+s-\sqrt{(a+b+s)^2-4ab}}{2a}\right)^n&\text{when}~(a+b+s)^2-4ab\neq0\\\Theta_1(n)n\left(\dfrac{a+b+s}{2a}\right)^n+\Theta_2(n)\left(\dfrac{a+b+s}{2a}\right)^n&\text{when}~(a+b+s)^2-4ab=0\end{cases}$ , where $\Theta_1(n)$ and $\Theta_2(n)$ are arbitrary periodic functions with unit period and $\begin{cases}\Theta_1(0)+\Theta_2(0)=1&\text{when}~(a+b+s)^2-4ab\neq0\\\Theta_2(0)=1&\text{when}~(a+b+s)^2-4ab=0\end{cases}$
Case $3$: $q\neq0$ and $q\neq1$
Then $(aq^n+b+s)C_n(s)=aq^nC_{n+1}(s)+bC_{n-1}(s)$
Let $C_n(s)=\int_{-\infty}^\infty q^{nt}K(t,s)~dt$ ,
Then $(aq^n+b+s)\int_{-\infty}^\infty q^{nt}K(t,s)~dt=aq^n\int_{-\infty}^\infty q^{(n+1)t}K(t,s)~dt+b\int_{-\infty}^\infty q^{(n-1)t}K(t,s)~dt$
$aq^n\int_{-\infty}^\infty q^{nt}K(t,s)~dt+(b+s)\int_{-\infty}^\infty q^{nt}K(t,s)~dt-aq^n\int_{-\infty}^\infty q^{(n+1)t}K(t,s)~dt-b\int_{-\infty}^\infty q^{(n-1)t}K(t,s)~dt=0$
$\int_{-\infty}^\infty aq^{n(t+1)}K(t,s)~dt+\int_{-\infty}^\infty(b+s)q^{nt}K(t,s)~dt-\int_{-\infty}^\infty aq^{n(t+1)}q^tK(t,s)~dt-\int_{-\infty}^\infty bq^{nt}q^{-t}K(t,s)~dt=0$
$\int_{-\infty}^\infty aq^{nt}K(t-1,s)~dt+\int_{-\infty}^\infty(b+s)q^{nt}K(t,s)~dt-\int_{-\infty}^\infty aq^{nt}q^{t-1}K(t-1,s)~dt-\int_{-\infty}^\infty bq^{nt}q^{-t}K(t,s)~dt=0$
$\int_{-\infty}^\infty((b+s-bq^{-t})K(t,s)+a(1-q^{t-1})K(t-1,s))q^{nt}~dt=0$
$\therefore(b+s-bq^{-t})K(t,s)+a(1-q^{t-1})K(t-1,s)=0$
$K(t,s)=-\dfrac{a(1-q^{t-1})K(t-1,s)}{b+s-bq^{-t}}$
$K(t,s)=\theta(t)(-1)^ta^t\prod\limits_{k=0}^\infty\dfrac{1-q^kq^t}{b+s-bq^{-k-1}q^{-t}}$ , where $\theta(t)$ is an arbitrary periodic functions with unit period
$\therefore C_n(s)=\Theta(n)\int_{-\infty}^\infty(-1)^ta^tq^{nt}\prod\limits_{k=0}^\infty\dfrac{1-q^kq^t}{b+s-bq^{-k-1}q^{-t}}dt$ , where $\Theta(n)$ is an arbitrary periodic functions with unit period
But this is only one of the group of the linear independent solution. I have no idea to find another group of the linear independent solution, since second order linear recurrence relations unlike second order linear differential equations which have reduction of order.
A: As
$$
G_n(s) = \frac{1}{a q^{n-1}}\left((a q^{n-1}+b+s)G_{n-1}(s)-b G_{n-2}(s)\right)
$$
with
$$
G_0(s) = 1, \ \ \ G_1(s) = \frac{c_1}{s+b_1}
$$
the recurrence formula which follows, in MATHEMATICA, generates all the instances for $G_n(s)$
G[s, 0] = 1;
G[s, 1] = Subscript[c, 1] /(s + b_0);
G[s_, n_] := ((a q^(n - 1) + b + s) G[s, n - 1] - b G[s, n - 2]) q^(1 - n)/a

G[s,4] // Expand

$$
G_4(s) = \frac{1}{s+b_1}\left(\frac{b^3 c_1}{a^3 q^6}-\frac{b^3 s}{a^3 q^6}-\frac{b_1 b^3}{a^3 q^6}+\frac{3 b^2 c_1 s}{a^3 q^6}-\frac{2 b^2 s^2}{a^3 q^6}-\frac{2
   b_1 b^2 s}{a^3 q^6}+\frac{3 b c_1 s^2}{a^3 q^6}-\frac{b s^3}{a^3 q^6}-\frac{b_1 b s^2}{a^3 q^6}+\frac{c_1 s^3}{a^3 q^6}+\frac{b^2
   c_1}{a^2 q^3}-\frac{b^2 s}{a^2 q^3}-\frac{b_1 b^2}{a^2 q^3}+\frac{b c_1 s}{a^2 q^5}+\frac{b c_1 s}{a^2 q^4}+\frac{2 b c_1 s}{a^2
   q^3}-\frac{b s^2}{a^2 q^4}-\frac{b_1 b s}{a^2 q^4}-\frac{b s^2}{a^2 q^3}-\frac{b_1 b s}{a^2 q^3}+\frac{c_1 s^2}{a^2 q^5}+\frac{c_1
   s^2}{a^2 q^4}+\frac{c_1 s^2}{a^2 q^3}+\frac{b c_1}{a q}-\frac{b s}{a q}-\frac{b_1 b}{a q}+\frac{c_1 s}{a q^3}+\frac{c_1 s}{a
   q^2}+\frac{c_1 s}{a q}+c_1\right)
$$
So $G_n(s)$ is non causal for $n \ge 2$.
