Functional equation $ (n+1) f(n+1)= (a n+b) f(n) $ for $n=0,1,...$ I am looking for a solution to the following functional equation:
\begin{align}
(n+1) f(n+1)= (a n+b) f(n), n=0,1,...
\end{align}
where $a$ and $b$ are some positive constants. 
Moreover, $f(n)$ is positive and 
\begin{align}
0<f(n) &\le 1,\\
\sum_{n=0}^\infty f(n)&=1. 
\end{align}
I was able to find a solution if $a=b \le 1$
\begin{align}
f(n)= \frac{c}{(1+c)^{n+1}}, \, c=\frac{1-a}{a}. 
\end{align}
My questions: 


*

*Is there a systematic way to solve this equation?

*Is the solution to this equation unique? 


Any reference would also be appreciated. 
 A: One way is to use generating functions. Define
\begin{align*}
g(x) = \sum_{n=0}^{\infty} f(n) x^n
\end{align*}
Henceforth, I will use the notation
\begin{align*}
g(x) \leftrightarrow\{f(n)\}
\end{align*}
to indicate that $g(x)$ has coefficients of $f(n)$ for $x^n$ in its series expansion. Then we have
\begin{align*}
g'(x) \leftrightarrow\{(n+1)f(n+1)\} 
\end{align*}
and
\begin{align*}
ax g'(x) + bg(x) \leftrightarrow \{(an+b)f(n)\}
\end{align*}
We can set equal the corresponding generating functions
\begin{align*}
g'(x) = axg'(x) + bg(x)
\end{align*}
which has solution
\begin{align*}
g(x) = c_1 (1 - ax)^{-b/a} = \sum_{n=0}^{\infty} c_1 (-a)^n \binom{-b/a}{n}x^n \leftrightarrow \left\{c_1 (-a)^n \binom{-b/a}{n}\right\}
\end{align*}
So,
\begin{align*}
f(n) = c_1 (-a)^n \binom{-b/a}{n}
\end{align*}
The constant $c_1$ can be found be evaluating
\begin{align*}
1 = \sum_{n=0}^{\infty} f(n) = g(1) = c_1 (1 - a)^{-b/a} \implies c_1 = (1 - a)^{b/a}
\end{align*}
So we find
\begin{align*}
f(n) = (1 - a)^{b/a} (-a)^n \binom{-b/a}{n}
\end{align*}
It's worth noting that $f(n)$ is always positive, so we also have $0 < f(n) \le 1$.
