15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS
that a particular solution of the formun=Aαnshould be tried. Substituting this
into (15.26) gives
Aαn+1=aAαn+kαn,from which it follows thatA=k/(α−a) and that there is a particular solution
having the formun=kαn/(α−a), providedα=a. For the special caseα=a,the
reader can readily verify that a particular solution of the formun=Anαnis appro-
priate. This mirrors the corresponding situation for linear differential equations
when the RHS of the differential equation is contained in the complementary
function of its LHS.
In summary, the general solution to (15.26) isun={
C 1 an+kαn/(α−a) α=a,C 2 an+knαn−^1 α=a,(15.27)withC 1 =u 0 −k/(α−a)andC 2 =u 0.
Second-order recurrence relationsWe consider next recurrence relations that involveun− 1 in the prescription for
un+1and treat the general case in which the intervening term,un, is also present.
A typical equation is thus
un+1=aun+bun− 1 +k. (15.28)As previously, the general solution of this isun=vn+wn,wherevnsatisfies
vn+1=avn+bvn− 1 (15.29)andwnisanyparticular solution of (15.28); the proof follows the same lines as
that given earlier.
We have already seen for a first-order recurrence relation that the solution tothe homogeneous equation is given by terms forming a geometric series, and we
consider a corresponding series of powers in the present case. Settingvn=Aλnin
(15.29) for someλ, as yet undetermined, gives the requirement thatλshould satisfy
Aλn+1=aAλn+bAλn−^1.Dividing through byAλn−^1 (assumed non-zero) shows thatλcould be either of
the roots,λ 1 andλ 2 ,of
λ^2 −aλ−b=0, (15.30)which is known as thecharacteristic equationof the recurrence relation.
That there are two possible series of terms of the formAλnis consistent with thefact that two initial values (boundary conditions) have to be provided before the
series can be calculated by repeated use of (15.28). These two values are sufficient
to determine the appropriate coefficientAfor each of the series. Since (15.29) is