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
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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explain this please

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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