# Reduce the 2nd order difference equation to 1st order

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?

• By the way, welcome to the site ! – Claude Leibovici Dec 12 '18 at 9:40

## 2 Answers

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, $$X_t=c_1 \alpha^t+c_2 (\alpha \beta)^t$$

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

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