Solving a Linear Constant Coefficient Difference Equation? I have the following difference equation:
$$y[n] + \alpha y[n-1] = \beta^nu[n]$$
with $-1 < \alpha < 1$ and $-1 < \beta < 1$. It is given  $y[n] = 0$ for $n < 0$ and $u[n]$ is the unit step function. 
I want to obtain the general solution.
My attempt:
homogeneous solution:
$$y_h[n] + \alpha y_h[n-1] = 0$$
proposal: $y_h[n] = Ar^n$ and substitute into the above
$$Ar^n + A\alpha r^{n-1} = 0 \to Ar^{n-1}(r+\alpha) = 0$$
or
$r = -\alpha$, so the homogeneous solution is as follows:
$$y_h[n] = A(-\alpha)^n$$
particular solution:
proposal: $y_p[n] = C\beta^n$ and using the governing equation:
$$C\beta^n + \alpha C \beta^{n-1} = \beta^n$$
which can be solved for C:
$$C = \frac{\beta^n}{\beta^n + \alpha\beta^{n-1}} = \frac{\beta}{\beta+\alpha}$$ 
which gives 
$$y_p[n] = \frac{\beta}{\beta+\alpha}\beta^n = \frac{\beta^{n+1}}{\beta + \alpha}$$
and the general solution:
$$y[n] = A(-\alpha)^n + \frac{\beta^{n+1}}{\beta + \alpha}u[n]$$
Question: 
Is the above approach correct? 
Especially the particular solution as I went from needing $\beta^n$ (current step) to $\beta^{n+1}$ (future step)
 A: For the homogeneous
$$
y_h^n+\alpha y_h^n = 0\Rightarrow y_h^n = (-\alpha)^n C_0
$$
now proposing for the particular
$$
y_p^n =  (-\alpha)^n C_n
$$
after substituting into the complete recurrence we get
$$
C_n-C_{n-1} = \alpha\left(-\frac {\beta}{\alpha}\right)^n\Rightarrow C_n=\beta\sum_{k=-1}^{n-1}\left(-\frac {\beta}{\alpha}\right)u_{k+1}
$$
and finally
$$
y_n = y_h^n+y_p^n =  (-\alpha)^n C_0 + (-\alpha)^n \left(\beta\sum_{k=-1}^{n-1}\left(-\frac {\beta}{\alpha}\right)^k u_{k+1}\right)
$$
A: Given
$$
y_{\,n}  + ay_{\,n - 1}  = b^{\,n} u(n)
$$
the z-transform is
$$
\eqalign{
  & \sum\limits_{0\, \le \,n} {\left( {y_{\,n}  + ay_{\,n - 1} } \right)z^{\,n} }  = \sum\limits_{0\, \le \,n} {b^{\,n} z^{\,n} }   \cr 
  & \sum\limits_{0\, \le \,n} {y_{\,n} z^{\,n} }  + az\sum\limits_{0\, \le \,n} {y_{\,n - 1} z^{\,n - 1} }  = \left( {1 + az} \right)F(z) = {1 \over {1 - bz}}  \cr 
  & F(z) = {1 \over {\left( {1 + az} \right)\left( {1 - bz} \right)}} = {a \over {a + b}}{1 \over {1 + az}} + {b \over {a + b}}{1 \over {1 - bz}} \cr} 
$$
and therefore the solution is
$$
y_{\,n}  = {a \over {a + b}}\left( { - a} \right)^{\,n}  + {b \over {a + b}}b^{\,n} 
$$
which you can verify is correct.
So your solution is also correct and you shall not get confused in writing it as 
$ .. b^{n+1}$ because $b/(a+b)$ is actually the "constant", since you need $y_0=1$.
