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I have the difference equation:

$x(n+1) = \beta + x(n)(1-\alpha - \beta)$, where $\alpha, \beta$ are constants, with initial condition $x(0) = 1$.

The solution says that the answer is $$x(n) = \frac{\beta}{\alpha + \beta} + \frac{\alpha}{\alpha + \beta}(1-\alpha - \beta)^n$$ but I'm getting $$x(n) = \frac{\beta}{\alpha + \beta} + (1-\alpha - \beta)^n$$ It claims that the $\dfrac{\alpha}{\alpha + \beta}$ is a general constant obtained from the initial conditions, but this is just $1$ so I don't know where I'm going wrong.

In particular, I start by solving the homogeneous equation $x(n+1) = x(n)(1-\alpha-\beta)$, which gets me $x(n)=x(0)(1-\alpha - \beta)^n = (1-\alpha-\beta)^n$, but I think that this may be wrong.

Working out:

Solve the homogeneous equation first - $x(n+1) = x(n)(1-\alpha - \beta)$.

Then, $x(n+1) = x(n)(1-\alpha-\beta) = x(n-1)(1-\alpha - \beta)(1 - \alpha - \beta) = x(n-1)(1-\alpha-\beta)^2 = \ldots = x(0)(1-\alpha-\beta)^n$

user341012
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    In your solution, $x(0) = {\beta \over \alpha + \beta} + 1 = {\alpha + 2 \beta \over \alpha + \beta} \neq 1$. How do you get $x(n) = x(0)(1-\alpha -\beta)^n$ and not $x(n) = C(1-\alpha -\beta)^n$ with $C$ being any constant at that point? – Abstraction May 20 '16 at 14:15
  • My solution is wrong indeed and I'm trying to figure out where the mistake is. I'll post my working out in an edit. – user341012 May 20 '16 at 14:19
  • You cannot just take the solution of the homogeneous equation as a solution of the inhomogeneous equation. Start by finding a solution to the inhomogeneous equation. You expect it to be a constant, so put $x(n)=\gamma$. Using the given relation you get $\gamma=\frac{\beta}{\alpha+\beta}$. Now the general solution is $x(n)=\frac{\beta}{\alpha+\beta}+C(1-\alpha-\beta)^n$. Use the initial condition to get $C$. – almagest May 20 '16 at 14:21
  • @almagest Ah, I forgot that you had to add a constant! Thank you! – user341012 May 20 '16 at 14:25

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Linear first-order difference equations of type $$y(n+1) = ay(n)+g(n),\ y(0)=y_0$$ have the following (unique) solution: $$y(n) = a^n y_0 + \sum_{k=0}^{n-1}a^{n-k-1}g(k) .$$ See An Introduction to Difference Equations by Elaydi. Applying this in your equation, you obtain $$x(n) = (1-\alpha-\beta)^n + \sum_{k=0}^{n-1} (1-\alpha-\beta)^{n-k-1}\beta.$$ It gives you $$x(n) = (1-\alpha-\beta)^n + \frac{\beta}{\alpha+\beta} - \frac{\beta}{\alpha+\beta}(1-\alpha-\beta)^n $$ and then $$ x(n)= \frac{\beta}{\alpha+\beta} + \beta\frac{\alpha+\beta-1}{\alpha+\beta}(1-\alpha-\beta)^n.$$

ACV
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