are klinearly independent solutions of the difference equation. To show this, solve
the difference equation
(E−a)kyn= 0, (10 .58)
as was done previously. Here the characteristic equation is (λ−a)k= 0 with λ=aas
a root of multiplicity k. The method of variation of parameters starts by assuming
a solution to equation (10.58) of the form yn=anUn. Observe that
Eyn=an+1Un+1 and (E−a)yn=an+1∆Un
E(E−a)yn=an+2∆Un+1 and (E−a)^2 yn=an+2∆^2 Un
E(E−a)^2 yn=an+3∆^2 Un+1 and (E−a)^3 yn=an+3∆^3 Un.
Continuing in this manner, show that Unmust satisfy
(E−a)kyn=an+k∆kUn= 0.
For a nonzero solution it is required that an+kbe different from zero, and so Unmust
be chosen such that ∆kUn= 0. The general solution of this equation is
Un=c 0 +c 1 n+c 2 n^2 +···+ck− 1 nk−^1 ,
and hence the general solution of equation (10.58) is yn=anUn.
One can compare the case of repeated roots for difference and differential equa-
tions and readily discern the analogies that exist.
Example 10-41. (Characteristic equation with complex or imaginary roots)
Solve the difference equation
yn+2 − 10 yn+1 + 74yn= (E^2 − 10 E+ 74)yn= 0.
Solution: Assume a solution of the form yn=λnand obtain the characteristic equa-
tion λ^2 − 10 λ+ 74 = 0 which has the complex roots λ 1 = 5 + 7 iand λ 2 = 6 − 7 i. Two
independent solutions are therefore
y 1 (n) = (5 + 7i)n and y 2 (n) = (6 − 7 i)n.
To obtain solutions in the form of real quantities, represent the roots λ 1 , λ 2 in
the polar form reiθ, as illustrated in figure 10-9.
