5.4.2 Example B Consider the equation z(k, l) + 2z(k – 1, l – 1) = l, (5.136) and let k – l = m. Therefore, setting で Uk = z(k, k – m) (5.137) ||
5.4.2 Example B Consider the equation z(k, l) + 2z(k – 1, l – 1) = l, (5.136) and let k – l = m. Therefore, setting で Uk = z(k, k – m) (5.137) ||
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 determine blue
Transcribed Image Text:5.4.2 Еxample B
Consider the equation
z(k, l) + 2z(k – 1, l – 1) = l,
(5.136)
and let k – l = m. Therefore, setting
z(k, k – m)
(5.137)
Uk =
LINEAR PARTIAL DIFFERENCE EQUATIONS
183
gives
Vk + 2vk–1 = k – m.
(5.138)
The solution to equation (5.138) is
Vk = C1(-2)* + (2/9 – 1/3m) + 1/3k.
(5.139)
If we now replace m by k-l and C1 by f(k-l), we obtain the general solution
to equation (5.136),
z(k, l) = (-2)* f(k – €) + 2/9+ /3l.
(5.140)
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