Example A The equation z(k+1, l) = 4z(k, l + 1) (5.32) can be written in symbolic form (E1 – 4E2)z(k, l) = 0. (5.33) Its solution is z(k, l) = (4E2)* A(e) = 4* E; A(t) (5.34) or C=(k, l) = 4* A(k + l). (5.35) Equation (5.32) is first order in both k and l, and its general solution contains one arbitrary function.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Example A
The equation
z(k +1, l) = 4z(k, l + 1)
(5.32)
can be written in symbolic form
(E1 – 4E2)z(k, l) = 0.
(5.33)
Its solution is
z(k, l) = (4E2)* A(l) = 4* E$ A(t)
(5.34)
or
C:(k, l) = 4* A(k + l).
(5.35)
Equation (5.32) is first order in both k and l, and its general solution contains
one arbitrary function.
Transcribed Image Text:Example A The equation z(k +1, l) = 4z(k, l + 1) (5.32) can be written in symbolic form (E1 – 4E2)z(k, l) = 0. (5.33) Its solution is z(k, l) = (4E2)* A(l) = 4* E$ A(t) (5.34) or C:(k, l) = 4* A(k + l). (5.35) Equation (5.32) is first order in both k and l, and its general solution contains one arbitrary function.
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