3.1.2 Example B The second-order difference equation Yk+2 – 3yk+1 + 2yk = 0 (3.24)

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Explain the determaine 

3.1.2 Example B
The second-order difference equation
Yk+2
3yk+1 + 2yk = 0
(3.24)
86
Difference Equations
(2)
has the two solutions y)
substitution into equation (3.24). For any two arbitrary constants, c1 and c2,
it is easily seen that
= 1 and y
2k, as can be shown by direct
(1)
(2)
C1Yk
+ c2Yk
= c1 + c22k
(3.25)
Yk =
is also a solution, i.e.,
(c1 + c22k+2) – 3(cı + c22*+1) + 2(c1+ c22k)
= (1 – 3+ 2)cı + (4 – 6 + 2)c22* = 0. (3.26)
Transcribed Image Text:3.1.2 Example B The second-order difference equation Yk+2 3yk+1 + 2yk = 0 (3.24) 86 Difference Equations (2) has the two solutions y) substitution into equation (3.24). For any two arbitrary constants, c1 and c2, it is easily seen that = 1 and y 2k, as can be shown by direct (1) (2) C1Yk + c2Yk = c1 + c22k (3.25) Yk = is also a solution, i.e., (c1 + c22k+2) – 3(cı + c22*+1) + 2(c1+ c22k) = (1 – 3+ 2)cı + (4 – 6 + 2)c22* = 0. (3.26)
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