184 Difference Equations Therefore, its solution is Vk = C1 + C2k + C3k², (5.150) and we conclude that the general solution of equation (5.146) is z(k, l) = f1(k + l) + k f2(k + l) + k² f3(k + l). (5.151)

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section4.6: Applications And The Perron-frobenius Theorem
Problem 70EQ
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5.4.4 Example D
The equation
z(k +3, l) – 32(k +2, l + 1) + 3z(k + 1, l + 2) – z(k, l + 3) = 0
(5.146)
has the sum of its arguments equal to a constant. Therefore, setting
k + l = m,
Vk = z(k, m – k),
(5.147)
we obtain
Vk+3 – 3vk+2 + 3vk+1
Uk =
= 0,
(5.148)
the characteristic equation of which is
p3 – 3r2 + 3r – 1 = (r – 1)3 = 0.
(5.149)
184
Difference Equations
Therefore, its solution is
Vk = C1 + C2k+ C3k²,
(5.150)
and we conclude that the general solution of equation (5.146) is
z(k, l) = f1(k + l) + k f2(k + l) + k² f3(k + l).
(5.151)
Transcribed Image Text:5.4.4 Example D The equation z(k +3, l) – 32(k +2, l + 1) + 3z(k + 1, l + 2) – z(k, l + 3) = 0 (5.146) has the sum of its arguments equal to a constant. Therefore, setting k + l = m, Vk = z(k, m – k), (5.147) we obtain Vk+3 – 3vk+2 + 3vk+1 Uk = = 0, (5.148) the characteristic equation of which is p3 – 3r2 + 3r – 1 = (r – 1)3 = 0. (5.149) 184 Difference Equations Therefore, its solution is Vk = C1 + C2k+ C3k², (5.150) and we conclude that the general solution of equation (5.146) is z(k, l) = f1(k + l) + k f2(k + l) + k² f3(k + l). (5.151)
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