Example H Consider the following second-order difference equation: Yk+2 – 2yk+1 + 2yk = 0. (4.58) The characteristic equation p2 – 2r + 2 = 0 (4.59) has the two solutions ri = 1+i, r2 = 1– i, (4.60) where i = V-1. These two complex conjugate roots can be put in the forms - V2e™i/4 = r3. (4.61) The corresponding complex conjugate set of fundamental solutions is (1) Yk 2k/2 e™ik/4 (2)* = Yk (4.62) %3D Using etio = cos 0 ± i sin 0, (4.63) and taking the appropriate linear combinations, we can form another set of fundamental solutions T = 2k/2 cos(Tk/4), = 2*/2 sin(tk/4). (4.64) Therefore, we finally obtain the general solution to equation (4.58); it is Yk = c12/2 cos(Tk/4) + c22*/2 sin(tk/4). (4.65)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Example H
Consider the following second-order difference equation:
Yk+2 – 2yk+1+ 2yk = 0.
(4.58)
The characteristic equation
p2 – 2r + 2 = 0
(4.59)
has the two solutions
ri = 1+ i, r2 = 1 – i,
(4.60)
where i = V–1. These two complex conjugate roots can be put in the forms
VZemi/4 = r3.
(4.61)
=
The corresponding complex conjugate set of fundamental solutions is
(1) – 2k/2 eTik/4 = y
(2)*
(4.62)
Using
etio
= cos 0 + i sin 0,
(4.63)
and taking the appropriate linear combinations, we can form another set of
fundamental solutions
2k/2 cos(Tk/4), T = 2*/2 sin(Tk/4).
(4.64)
Therefore, we finally obtain the general solution to equation (4.58); it is
Yk =
c12*/2 cos(Tk/4) + c22*/² sin(rk/4).
(4.65)
Transcribed Image Text:Example H Consider the following second-order difference equation: Yk+2 – 2yk+1+ 2yk = 0. (4.58) The characteristic equation p2 – 2r + 2 = 0 (4.59) has the two solutions ri = 1+ i, r2 = 1 – i, (4.60) where i = V–1. These two complex conjugate roots can be put in the forms VZemi/4 = r3. (4.61) = The corresponding complex conjugate set of fundamental solutions is (1) – 2k/2 eTik/4 = y (2)* (4.62) Using etio = cos 0 + i sin 0, (4.63) and taking the appropriate linear combinations, we can form another set of fundamental solutions 2k/2 cos(Tk/4), T = 2*/2 sin(Tk/4). (4.64) Therefore, we finally obtain the general solution to equation (4.58); it is Yk = c12*/2 cos(Tk/4) + c22*/² sin(rk/4). (4.65)
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