First I would rewrite the recursion using
$C_k = B_{n-1-k}$
s.t. $C_0$ is the base case and we get the recursion
$C_k = 1 + \frac{k}{n}C_{k-1} + \frac{n-k-1}{n}x, \qquad k>0.$
Next one can split $C_k = \frac{D_k}{n^k} + x\frac{E_k}{n^{k+1}}$ into a term with $x$ and a term without $x$, where the denominators $n^k, n^{k+1}$ are for convenience. This gives the recursions
$D_k = kD_{k-1} + n^k, \quad k>0, \quad D_0 = 1$
and
$E_k = k E_{k-1} + (n-k-1)n^k, \quad k>0, \quad E_0=n-1.$
Both recursions can be expanded into simple sums. Thus we get
$C_k = \frac1{n^k}\sum_{i=0}^k (k)_i n^{k-i} + \frac{x}{n^{k+1}}\sum_0^k \mbox{something similar}$
where $(k)_i = k(k-1)\ldots (k-i+1)$.
(I hope I got all the indices and exponents right in my quick writing; but the principle should definitely work.)