Now consider the fourth-order equation Yk+4 + Yk = 0. (4.76) Its characteristic equation is G+1= (* + )(r² – i) = 0. (4.77) From the first factor, we obtain ri = e-in/4 r2 = e -3in/4 (4.78) and from the second factor r3 = e -in/4 -Зіт /4 (4.79) Since ri = r and r2 = ri, we have the following four fundamental solutions: r4 = e

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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Difference Equations
Example J
Now consider the fourth-order equation
Yk+4 + Yk = 0.
(4.76)
Its characteristic equation is
G+1=
( +1 = (r² + i)(p² – i) = 0.
(4.77)
From the first factor, we obtain
ri = e-in/4
- 3 ίπ/4
(4.78)
r2 = e
and from the second factor
-in /4.
-3it/4.
(4.79)
r3 = e
r4 = e
Since ri = r* and r2 = ri, we have the following four fundamental solutions:
y = cos(Tk/4),
= sin(Tk/4),
(4.80)
(3)
(4)
cos(3tk/4), Y
= sin(37k/4).
Consequently, the general solution to equation (4.71) is
Yk = C1 cos(Tk/4)+ c2 sin(rk/4)
+ Сз COs(3пk/4) + с4 sin(3rk/4),
(4.81)
where c1, c2, C3, and c4 are arbitrary constants.
Transcribed Image Text:Difference Equations Example J Now consider the fourth-order equation Yk+4 + Yk = 0. (4.76) Its characteristic equation is G+1= ( +1 = (r² + i)(p² – i) = 0. (4.77) From the first factor, we obtain ri = e-in/4 - 3 ίπ/4 (4.78) r2 = e and from the second factor -in /4. -3it/4. (4.79) r3 = e r4 = e Since ri = r* and r2 = ri, we have the following four fundamental solutions: y = cos(Tk/4), = sin(Tk/4), (4.80) (3) (4) cos(3tk/4), Y = sin(37k/4). Consequently, the general solution to equation (4.71) is Yk = C1 cos(Tk/4)+ c2 sin(rk/4) + Сз COs(3пk/4) + с4 sin(3rk/4), (4.81) where c1, c2, C3, and c4 are arbitrary constants.
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