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The elimination of ci and c2 gives the relation Yk 1 1 2 0, (1.53) Yk+1 4 Yk+2 25 which, on expanding, is equal to Yk+2 - 7yk+1 + 10yk = 0. (1.54)

The elimination of ci and c2 gives the relation Yk 1 1 2 0, (1.53) Yk+1 4 Yk+2 25 which, on expanding, is equal to Yk+2 - 7yk+1 + 10yk = 0. (1.54)

Advanced Engineering Mathematics
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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1.3.2 Example B Let yk be given by the expression Yk = c12* + c25*, (1.51) where ci and c2 are arbitrary constants. Thus, Yk must satisfy a second-order difference equation. To determine this equation, we calculate yk+1 and yk+2: Yk+1 = 2c12* + 5c25*, Yk+2 = 4c12* + 25c25k. (1.52) THE DIFFERENCE CALCULUS The elimination of c1 and c2 gives the relation Yk 1 1 0, (1.53) Yk+1 4 %| Yk+2 25 which, on expanding, is equal to Yk+2 – 7yk+1 + 10yk 0. (1.54)
Transcribed Image Text:1.3.2 Example B Let yk be given by the expression Yk = c12* + c25*, (1.51) where ci and c2 are arbitrary constants. Thus, Yk must satisfy a second-order difference equation. To determine this equation, we calculate yk+1 and yk+2: Yk+1 = 2c12* + 5c25*, Yk+2 = 4c12* + 25c25k. (1.52) THE DIFFERENCE CALCULUS The elimination of c1 and c2 gives the relation Yk 1 1 0, (1.53) Yk+1 4 %| Yk+2 25 which, on expanding, is equal to Yk+2 – 7yk+1 + 10yk 0. (1.54)
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