9.2 • SECOND-ORDER HOMOGENEOUS DIFFERENCE EQUATIONS 363Thus the solution isy[n] = Res[f(z), 1] + Res[f(z), 3)
= -1+2. 3»,
which agrees with the result of Example 9. 12.(b) Take the z-transforms of each term.z^2 (Y(z)-1- 3z-^1 ) - 4(z(Y(z)-1)) + 3(Y(z)) = z ~
2.S 1 £ Y( ) d ( ) z(^3) - 3z (^2) + 6z
o ve or z an get Y z = (•-l)(z- 2 )(%-3)"
Calculate the residues for f (z) = Y(z)zn- l = z
3
-^3 •
2
(z-l)(z-2)(z±^6 • -3) zn-1 at
the poles
[ ( ) I. z
(^3) - 3z (^2) + 6z n-I n -I
Res f z , 1 = z-1 hrn ( z. - 2 )( Z - 3 ) z = 2 · 1 = 2, and
= lim z3 - 3z2 + 6z zn-1 = - 8. 2n-1 = -2n+2 and
z- 2 (z - l)(z - 3) •
Res[f(.z), 3] = lim z3 - 3z2 + 6z = 9. 3n-I = 3n±1 _
z- 3 (z - l)(z - 2)
Thus the solution isy{n] = Res[f(z), 1) + Res[f(z), 2) + Res[f(z), 3]
= 2 - 2n+2 + 3n±1.- EXAMPLE 9.16
(a) Use z-transforrn methods to solve y[n + 2) - 4y[n + l ] + 4y(n) = 0 with
y(O] = Yo = 1 and y (l] =YI = 5.
(b) Use z-transforrn methods to solve y(n + 2] - 4y(n + 1) + 4y[n) = 3n
wit h y[OJ = Yo = 2 and y(l] = YI = 3.
Solution
(a) Take the z-transforms of each termz^2 (Y(z) - 1 - 5.z-^1 ) - 4(z(Y(z) - 1)) + 4(Y (z)) = O.
Solve for Y(z) and get Y(z) = (~~"t)2·