Exercise 7.3.5
Let $\lambda_1$ and $\lambda_2$ be the roots of the characteristic equation of the second order-difference equation
$$x(n+2)+\alpha x(n+1)+\beta x(n)=0. \quad (7.4.9)$$
(a) Write (7.4.9) using the operator E and get
$$(E^2+\alpha E+\beta)x(n)=0.$$
(b) Show that
$$E^2+\alpha E+\beta=(E-\lambda_1)(E-\lambda_2).$$
(c) In this notation, show that the difference equation (7.4.9) takes the form
$$(E-\lambda_1)(E-\lambda_2)x(n). \quad (7.4.10)$$
(d) Let $y(n)$ be the solution of $(E-\lambda_2)x(n)=0$. In other words,
$$(E-\lambda_2)x(n)=y(n).$$
Then (7.4.10) becomes
$$(E-\lambda_1)y(n). \quad (7.4.11)$$
(e) Show that (7.4.11) has the solution $y(n)=\lambda_1^n$.
(f) Show that with this $y(n)$, the difference equation in part (d) reduces to
$$x(n+1)-\lambda_2 x(n)=\lambda_1^n. \quad (7.4.12)$$
(g) Refer to Section 7.3.1 to show that solution of (7.4.12) is
$$x(n)=\begin{cases} c_1\lambda_1^n+c_2\lambda_2^n, & \text{for } \lambda_1 \neq \lambda_2 \\ c_1\lambda_1^n+c_2n\lambda_1^n, & \text{for } \lambda_1 = \lambda_2. \end{cases}$$
