1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

9.2 • SECOND-ORDER HOMOGENEOUS D IFFERENCE EQUATIONS 359

9.2.1 Remark About Stability

Stability depends on the location of the roots of the characteristic polynomial.
Without loss of generality, if lr 1 I :5: h i < 1, then both roots lie inside the unit
circle and the solutions are asymptotically stable and tend to zero as n -+ oo. If
r 1 =I r2 and h i :5: hi= 1, then a root lies on the unit circle and the solutions
are stable. If r 1 = r2 = ±1, then there is an unstable solution. Finally, if 1 < hi,
then at least one root lies outside the unit circle and there is an unstable solution.

• EXAMPLE 9.12 Solve y[n+ 2] - 4y[n+ 1) + 3y[n] := 0 with y[O) = y 0 = 1 and

y(l] = YI= 5.

Solution The characteristic equation r^2 - 4r+3 = (r - l)(r- 3) = 0 has roots
r 1 = 1 and r2 = 3, hence the general solution is y[n) = c1 + c 23 ". Use the initial
conditiollS and form the linear system

y[O] = C1 + C2. 1 = 1 and y[l) = C1 + C2. 3 = 5,

then solve for the constants and get c 1 = -1, c 2 = 2. Hence the solution is

y[n) = - 1 + 2 · 3".

• EXAMPLE 9.13 Solve y\n + 2) -4y[n + l] + 4y[n) = 0 with y[O] = y 0 = 1 and

y(l ] =YI= 5.

Solution The characteristic equation 1·^2 - 4r + 4 = (r - 2)^2 = 0 has equal

roots r = 1· 1 = r2 = 2, hence the general solution is y[n] = c 1 2" + c2n 2". Use

the initial conditiollS and form the linear system

y[O) =CI + Cz. 0 = 1 and y(l] = CJ. 2 + Cz. 2 = 5,

then solve for the constants and get CI = 1 and ez = ~· Hence the solution is

y[n) = 2" + ~n 2" = 2" + 3n · 2n-t.

2

•EXAMPLE 9.14 Solvey(n+2]- 4y[n+1)+5y[n) = 0 with y[O] =Yo= 1 and

y (l] =YI= 5.

Solution The characteristic equation r^2 - 4r + 5 = (r - 2 + i)(r - 2 - i) = 0

bas complex roots r 1 = 2 + i and r2 = 2 - i , hence the general solut ion is