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)
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 $y_0=1$.