5.2.6 Example F Consider the first-order linear partial difference equation z(k +1, l+ 1) – kz(k, l) = 0. (5.59) Note that the coefficient of the z(k, l) term is not constant, but is equal to k. In operator form, this equation becomes (E¡E2 – k)z(k, l) = 0, (5.60) or (E1 – kE,')z(k, l) = 0. (5.61) | In terms of E1, this equation is a first-order linear equation whose solution is k-1 ( 2(k, e) = II (iE,')«(e) = (k – 1)!Ez*+1 ø(0). (k, l): $(C). (5.62) %3D i=1 Therefore, the general solution to equation (5.59) is z(k, l) = (k – 1)!ø(e – k + 1). (5.63)
5.2.6 Example F Consider the first-order linear partial difference equation z(k +1, l+ 1) – kz(k, l) = 0. (5.59) Note that the coefficient of the z(k, l) term is not constant, but is equal to k. In operator form, this equation becomes (E¡E2 – k)z(k, l) = 0, (5.60) or (E1 – kE,')z(k, l) = 0. (5.61) | In terms of E1, this equation is a first-order linear equation whose solution is k-1 ( 2(k, e) = II (iE,')«(e) = (k – 1)!Ez*+1 ø(0). (k, l): $(C). (5.62) %3D i=1 Therefore, the general solution to equation (5.59) is z(k, l) = (k – 1)!ø(e – k + 1). (5.63)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:5.2.6 Example F
Consider the first-order linear partial difference equation
z(k +1, l+ 1) – kz(k, l) = 0.
(5.59)
Note that the coefficient of the z(k, l) term is not constant, but is equal to k.
In operator form, this equation becomes
(E,E2 – k)z(k, l) = 0,
(5.60)
or
(E1 – kE,')z(k, l) = 0.
(5.61)
In terms of E1, this equation is a first-order linear equation whose solution is
k-1
(2(k, l) = II (iE,')¢(e) = (k – 1)!E,k+1 6(e).
(5.62)
i=1
Therefore, the general solution to equation (5.59) is
z(k, l) = (k – 1)!$(l – k + 1).
(5.63)
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