Example 10-40. (Characteristic equation with repeated roots)
Find the general solution to the difference equation
yn+2 − 4 yn+1 + 4yn= 0.Solution: Write the difference equation in operator form (E^2 − 4 E+ 4)yn = 0 and
assume a solution of the form yn=λn. By substituting the assumed solution into
the difference equation one obtains the characteristic equation λ^2 − 4 λ+ 4 = 0 which
has the repeated roots λ= 2 , 2. As with ordinary differential equations, one solution
is y 1 (n) = 2nand the second independent solution can be obtained by a multiplication
of the first solution by the independent variable n. This is analogous to the case
of repeated roots for ordinary differential equations with constant coefficients. A
second independent solution is therefore y 2 (n) = n 2 n,and the general solution can be
expressed as
y(n) = yn=c 02 n+c 1 n 2 n,where c 0 and c 1 are arbitrary constants. To verify that n 2 n is a second indepen-
dent solution, the method of variation of parameters is used. Assume that a second
solution has the form yn=Un 2 n,where Un is an unknown function of nto be deter-
mined. Substituting this assumed solution into the difference equation produces the
equation
2 n+2(Un+2 − 2 Un+1 +Un) = 0which can be written as
(E^2 − 2 E+ 1)Un= (E−1)^2 Un= ∆^2 Un= 0. (10 .57)It is left as an exercise to verify that the general solution of ∆kUn= 0 is given by
Un=c 0 +c 1 n+c 2 n^2 +c 3 n^3 +···+ck− 1 nk−^1 ,and therefore, equation (10.57) has the solution Un=c 0 +c 1 n, which when substituted
into the assumed solution gives the result y(n) = c 0 (2)n+c 1 n(2)nas the general solution.
In general, if the characteristic equation associated with a linear difference equa-
tion with constant coefficients has a characteristic root λ=aof multiplicity k, then
y 1 (n) = an, y 2 (n) = na n, y 3 (n) = n^2 an,... , y k(n) = nk−^1 an