Yk = B1(k) – 1/½B2(k) + /½B4(k) + A1,

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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FIRST-ORDER DIFFERENCE EQUATIONS
53
2.3.1
Example
The equation
Yk+1 – Yk = 1 - k+ 2k3
(2.62)
has the particular solution
k-1
k-1
k-1
k-1
Σ1-i+2) - Σ (1) -Σι+2Σε
Yk =
i=1
(2.63)
i=1
i=1
i=1
k(k – 1)
(k – 1)²k²
= (k – 1) –
2
The general solution is
Yk = /2k* – k3 + 3/2k + A,
(2.64)
where A is an arbitrary constant. In terms of Bernoulli polynomials, this last
expression reads
Yk = B1(k) – 1/2B2(k)+\/½B4(k) + A1,
(2.65)
where A1 is an arbitrary function.
Transcribed Image Text:FIRST-ORDER DIFFERENCE EQUATIONS 53 2.3.1 Example The equation Yk+1 – Yk = 1 - k+ 2k3 (2.62) has the particular solution k-1 k-1 k-1 k-1 Σ1-i+2) - Σ (1) -Σι+2Σε Yk = i=1 (2.63) i=1 i=1 i=1 k(k – 1) (k – 1)²k² = (k – 1) – 2 The general solution is Yk = /2k* – k3 + 3/2k + A, (2.64) where A is an arbitrary constant. In terms of Bernoulli polynomials, this last expression reads Yk = B1(k) – 1/2B2(k)+\/½B4(k) + A1, (2.65) where A1 is an arbitrary function.
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