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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Show me the steps of determine blue

5.5.2 Example B
Consider the equation
z(k +1, l + 1) + 22(k, l) = k² + l + 1,
(5.170)
or
(E1E2 + 2)z(k, e) = k² + l + 1.
(5.171)
The particular solution is
Zp(k, l) =
1
(k² +l+1)
E1 E2 + 2
1
1
- (k² + l + 1)
31+ 1/3(A1 + A2 + A¡A2)
= /3[1-/3(A1 + A2 + A1A2)
+ 1/(A? + Až + AA? + 2A¡A2 + 2A{A2
+2A¡A})+ · · ·](k² +l+1)
= 1/27(9k2 – 6k + 9l + 5).
(5.172)
The homogeneous equation
(E1, E2 + 2)zh(k, l) = 0
(5.173)
has the solution
zn (k, l) = (-2)*f(l – k).
(5.174)
Therefore, the general solution of equation (5.170) is
Zn(k, l) = zh(k, l) + žp(k, l)
(5.175)
= (-2)* f(l – k) + 1/27(9k² – 6k + 9l + 5).
Transcribed Image Text:5.5.2 Example B Consider the equation z(k +1, l + 1) + 22(k, l) = k² + l + 1, (5.170) or (E1E2 + 2)z(k, e) = k² + l + 1. (5.171) The particular solution is Zp(k, l) = 1 (k² +l+1) E1 E2 + 2 1 1 - (k² + l + 1) 31+ 1/3(A1 + A2 + A¡A2) = /3[1-/3(A1 + A2 + A1A2) + 1/(A? + Až + AA? + 2A¡A2 + 2A{A2 +2A¡A})+ · · ·](k² +l+1) = 1/27(9k2 – 6k + 9l + 5). (5.172) The homogeneous equation (E1, E2 + 2)zh(k, l) = 0 (5.173) has the solution zn (k, l) = (-2)*f(l – k). (5.174) Therefore, the general solution of equation (5.170) is Zn(k, l) = zh(k, l) + žp(k, l) (5.175) = (-2)* f(l – k) + 1/27(9k² – 6k + 9l + 5).
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