There appears to be a contradiction with the recurrence solutiom
From the comments under Tim Wescott's answer, it seems that there is still  some confusion concerning the correct way to solve such problems. Let's consider the non-homogeneous difference equation (DE)
$$y[n]-p y[n-1]=q^n,\qquad p\neq q\tag{1}$$
with initial value $y[-1]=c$. Note that initially we don't restrict the forcing function on the right-hand side of $(1)$ to equal zero for $n&lt;0$.
As correctly shown in the OP, the general solution to the homogeneous DE has the form
$$y_h[n]=Ap^n\tag{2}$$
which is also called the natural response,
and a particular solution (forced response) of $(1)$ is given by
$$y_p[n]=Bq^n\tag{3}$$
The constant $B$ is determined by substituting $(3)$ into $(1)$:
$$Bq^n-pBq^{n-1}=q^n\tag{4}$$
from which we obtain $B=\frac{q}{q-p}$. Consequently, the general solution of the non-homogeneous DE $(1)$ is
$$y[n]=y_h[n]+y_p[n]=Ap^n+\frac{q}{q-p}q^n\tag{5}$$
Note that the solution $(5)$ is valid for all values of $n$.
The value of $A$ is determined by the initial condition $y[-1]=c$. Substituting $n=-1$ into $(5)$ and setting it equal to $c$ results in
$$A=cp-\frac{p}{q-p}\tag{6}$$
The solution to the initial value problem is hence given by
$$y[n]=cp^{n+1}+\frac{1}{q-p}\left[q^{n+1}-p^{n+1}\right]\tag{7}$$
We now only consider values $n\ge 0$, simply because that's the range we're interested in.
Note that the term $cp^{n+1}$ in $(7)$ is the zero-input response (ZIR), and the remaining term is the zero-state response (ZSR).
From an engineering point of view, composing the solution into a ZSR and a ZIR appears more natural and more useful than the classical decomposition into a homogeneous and a particular solution, because the decomposition of the solution into ZSR and ZIR separates the contributions arising from an external input and from internal conditions, respectively.