Solution: In the given difference equation, replace nby n+ 1 in all terms, to obtain
yn+2 =yn+1 + 2yn,
then one can verify
n= 0, y 2 =y 1 + 2y 0 = 2
n= 1, y 3 =y 2 + 2y 1 = 2
n= 2, y 4 =y 3 + 2y 2 = 6
n= 3, y 5 =y 4 + 2y 3 = 10
n= 4, y 6 =y 5 + 2y 4 = 22
n= 5, y 7 =y 6 + 2y 5 = 42
n= 6, y 8 =y 7 + 2y 6 = 86
n= 7, y 9 = 7 8 + 2y 7 = 170
n= 8, y 10 =y 9 + 2 y 8 = 342.
The study of difference equations with constant coefficients closely parallels the
development of ordinary differential equations. Our goal is to determine functions
yn=y(n),defined over a set of values of n, which reduce the given difference equation
to an identity. Such functions are called solutions of the difference equation. For
example, the function yn= 3 nis a solution of the difference equation yn+1 − 3 yn= 0
because 3 n+1− 3 · 3 n= 0 for all n= 0, 1 , 2 ,.. .. Recall that for linear differential equations
with constant coefficients one can assume a solution of the form y(x) = exp(ωx ). This
assumption leads to producing the characteristic equation and consequently the
characteristic roots associated with the differential equation. In the special case
x=n, there results y(n) = yn= exp(ωn ) = λn,where λ= exp(ω) is a constant. This
suggests in our study of difference equations with constant coefficients that one
should assume a solution of the form yn=λn, where λis a constant. Analogous to
ordinary linear differential equations with constant coefficients, a linear, nth-order,
homogeneous difference equation with constant coefficients has associated with it
a characteristic equation with characteristic roots λ 1 , λ 2 ,... , λ n. The characteristic
equation is found by assuming a solution yn=λn, where λis a constant. The various
cases that can arise are illustrated by the following examples.
