 
Summary: Linear Difference Equations
Posted for Math 635, Spring 2012.
Consider the following secondorder linear difference equation
f(n) = af(n  1) + bf(n + 1), K < n < N, (1)
where f(n) is a function defined on the integers K n N, the value N can be chosen to be
infinity, and a and b are nonzero real numbers. Note that if f satisfies (1) and if the values f(K),
f(K + 1) are known then f(n) can be determined for all K n N recursively via the formula
f(n + 1) =
1
b
f(n)  af(n  1) .
Note also that if f1 and f2 are two solutions of (1), then c1f1 + c2f2 is a solution for any real
numbers c1, c2. Therefore, the solution space of (1) is a twodimensional vector space and one basis
for the space is {f1, f2} with f1(K) = 1, f1(K + 1) = 0 and f2(K) = 0, f2(K + 1) = 1.
We will solve (1) by looking for solutions of the form f(n) = n, for some = 0. Plugging n
into equation (1) yields
n
= an1
+ bn+1
, K < n < N,
