Reduce the 2nd order difference equation to 1st order

Question

I have the following 2nd order difference equation.

$\alpha X_{t+1}-X_{t+2} = \beta\alpha \left(\alpha X_{t}-X_{t+1}\right)$

Clearly, one solution is the process of $\alpha X_{t}=X_{t+1}$.

However, there is another solution which is $X_{t+1}=\beta\alpha X_{t}$.

The question is, WITHOUT the guess-and-verify method, how to derive the second one from the 2nd order difference equation at the beginning?

Answer 1

Calling

$$ Y_t = \alpha X_t-X_{t+1} $$

follows

$$ Y_{t+1} = \beta\alpha Y_t $$

so solving first for $Y_t$ we have

$$ Y_t = C_1(\alpha\beta)^{t-1} $$

and finally solving

$$ \alpha X_t-X_{t+1} = C_1(\alpha\beta)^{t-1} $$

we get

$$ X_t = C_2 \alpha ^{t-1}+\frac{C_1 \left(\alpha ^t-(\alpha \beta )^t\right)}{\alpha ^2 (\beta -1) \beta } $$

Answer 2

If you consider the recurrence equation $$\alpha X_{t+1}-X_{t+2} = \alpha \beta \left(\alpha X_{t}-X_{t+1}\right)$$ let $X_t=e^{kt}$, replace, simplify to get $$-\alpha^2 \beta+\alpha (\beta+1) e^k-e^{2 k}=0$$ and solve for $e^k$.

You should find two roots corresponding to $k=\log(a)$ and $k=\log(a b)$ making, as a general solution, $$X_t=c_1 \alpha^t+c_2 (\alpha \beta)^t$$