Example E Let us determine the second-order difference equation having the fundamen system (1) = k!, (2) Yk (4.10 = 1. We have Yk k! 1 Yk 1 1 k! Yk+1 (k + 1)! 1 Yk+2 (k + 2)! Yk+1 k +1 1 0, (4.10 Yk+2 (k +2)(k + 1) 1 which gives kyk+2 – (k² + 3k + 1)yk+1 + (k + 1)²yk = 0. (4.10
Example E Let us determine the second-order difference equation having the fundamen system (1) = k!, (2) Yk (4.10 = 1. We have Yk k! 1 Yk 1 1 k! Yk+1 (k + 1)! 1 Yk+2 (k + 2)! Yk+1 k +1 1 0, (4.10 Yk+2 (k +2)(k + 1) 1 which gives kyk+2 – (k² + 3k + 1)yk+1 + (k + 1)²yk = 0. (4.10
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Explain the determine
Transcribed Image Text:Example E
Let us determine the second-order difference equation having the fundamen
system
(1)
(2)
= k!,
= 1.
(4.10
We have
Yk
k!
1
Yk
1
1
(k + 1)! 1
Yk+2 (k + 2)!
k +1
(k +2)(k + 1)
Yk+1
k! Yk+1
1
0,
(4.10
Yk+2
1
which gives
kyk+2 – (k² + 3k + 1)yk+1 + (k +1)²yk =
0.
(4.10
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