The equation Yk+1 – Yk = 1 - k+ 2k3 has the particular solution k-1 k-1 k-1 k-1 J:-Σα-i+ 26) Σ(1) -Σί+2Σ %3D %3D i=1 i=1 i=1 i=1 (k – 1)²k² ,2 k(k – 1) | | = (k – 1) – - The general solution is 1o14 13 3ok
The equation Yk+1 – Yk = 1 - k+ 2k3 has the particular solution k-1 k-1 k-1 k-1 J:-Σα-i+ 26) Σ(1) -Σί+2Σ %3D %3D i=1 i=1 i=1 i=1 (k – 1)²k² ,2 k(k – 1) | | = (k – 1) – - The general solution is 1o14 13 3ok
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 detemaine step by step
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
υ-Σ1-ί+ 28) Σ (1) - Σί+2Σ
i=1
(2.63)
i=1
i=1
i=1
= (k – 1)
k(k – 1)
+
(k – 1)²k²
2
The general solution is
Yk = 1/½k4 – k³ + 3/2k + A,
(2.64)
where A is an arbitrary constant. In terms of Bernoulli polynomials, this last
expression reads
Yk
B1(k) – 1/½B2(k) + /½B4(k) + A1,
(2.65)
where A1 is an arbitrary function.
The result of equation (2.65) could have been immediately written down
by first noting that equation (2.60) is a linear equation and thus its particular
solution is a sum of particular solutions of equations having the form
Yk+1 –
Yk = amk", 0<m<n.
(2.66)
Under the transformation
Am
Вт+1(k),
т +1
(2.67)
Yk
equation (2.66) becomes
Bm+1(k+1) – Bm+1(k) = (m + 1)k",
(2.68)
which is the defining difference equation for the Bernoulli polynomials. There-
fore, from the known coefficients (am), the particular solution can be imme-
diately obtained from equation (2.67).
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