I'm solving the following recurrence:
Find the closed form of the recurrence relation $$ \begin{cases} x_{n+1} = {x_n \over b + x_n} \\ x_1 = a \end{cases} $$ Where $a, b$ are some constants.
I've started with a substitution:
$$ z_n = {1 \over x_n} $$
Rewrite to obtain the form of characteristic equation: $$ z_{n+1} = bz_n + 1 $$
Solve for homogenous and particular parts: $z_n = z_n^{(h)} + z_n^{(p)}$
$$ z_n^{(h)} = C\cdot b^n \\ z_n^{(p)} = {1\over 1 -b} $$
Particular is obtained by assuming that particular solution is some constant $B$. Now using the initial conditions $x_1 = a$ and doing reverse substitution:
$$ \begin{align} a &= \frac{1-b}{C\cdot b^n(1-b) + 1} \iff \\ \iff C &= \frac{1-b-a}{ab(1-b)} \end{align} $$
Which gives the closed form for $x_n$:
$$ x_n = \frac{(1-b)^2ab}{(1-b)((1-b-a)b^n+ab)} $$
So here if we assume that $b \ne 1$ then $x_n$:
$$ \begin{cases} x_n = \frac{a(b-1)}{(a+b-1)b^{n-1} - a} \\ b \ne 1 \\ a \ne \frac{-(b-1)b^k}{b^k - 1} \\ k, n \in \mathbb N \\ n \ge 2 \end{cases} $$
This matches the answer from the keys section. However keys suggest yet another solution assuming $b = 1$ and $a \ne -{1\over k}$. That solution is as follows:
$$ \begin{cases} x_n = \frac{a}{1+(n-1)a} \\ b = 1 \\ a \ne -{1 \over k} \\ k, n \in \mathbb N \end{cases} $$
I don't really understand where the second solution comes from. How can we assume that $b = 1$? Won't it give a nonsense implying division by zero in the general term for $x_n$?
$b=1$ gives a perfectly good recurrence relation $x_{n+1}=\dfrac{x_n}{1+x_n}$ for almost all $a$. You assumed $b\neq 1$ in order to impose the guess $z_n^{(p)}=\frac{1}{b-1}$.
Indeed, the recurrence relation for $z_n$ becomes very easy to solve when $b=1$: it is just adding $1$ at every step so $z_n=z_1+n-1$.