Linear Difference Equations Posted for Math 635, Spring 2012. Summary: Linear Difference Equations Posted for Math 635, Spring 2012. Consider the following second-order 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 two-dimensional 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 = an-1 + bn+1 , K < n < N, Collections: Mathematics