5.5.2 Еxample B Consider the equation z(k + 1, l + 1) + 2z(k, e) = k2 + l +1, (5.170) or (E, E2 + 2)z(k, l) = k² + l+ 1. (5.171) The particular solution is Zp(k, l) 1 (k² +l+1) E, E2 + 2 1 1 -(k² +l+1) 31+1/3(A1 + A2 + A¡A2) = 1/3[1-1/3(A1 + A2 + A¡A2) +1/o(△? + △2 + Δ우스2 + 2△1△2 + 2A{△2 +2Δ1Δ2)+.](k2 + l+ 1) = 1/27(9k? – 6k + 9l + 5). (5.172) The homogeneous equation (E1, E2 + 2)ž½(k, l) = 0 (5.173) has the solution zh(k, l) = (-2)* f (e – k). (5.174) Therefore, the general solution of equation (5.170) is Zh(k, l) = zh(k, e) + zp(k, l) (5.175) = (-2)* f(l – k) + /27(9k² – 6k + 9l + 5).
5.5.2 Еxample B Consider the equation z(k + 1, l + 1) + 2z(k, e) = k2 + l +1, (5.170) or (E, E2 + 2)z(k, l) = k² + l+ 1. (5.171) The particular solution is Zp(k, l) 1 (k² +l+1) E, E2 + 2 1 1 -(k² +l+1) 31+1/3(A1 + A2 + A¡A2) = 1/3[1-1/3(A1 + A2 + A¡A2) +1/o(△? + △2 + Δ우스2 + 2△1△2 + 2A{△2 +2Δ1Δ2)+.](k2 + l+ 1) = 1/27(9k? – 6k + 9l + 5). (5.172) The homogeneous equation (E1, E2 + 2)ž½(k, l) = 0 (5.173) has the solution zh(k, l) = (-2)* f (e – k). (5.174) Therefore, the general solution of equation (5.170) is Zh(k, l) = zh(k, e) + zp(k, l) (5.175) = (-2)* f(l – k) + /27(9k² – 6k + 9l + 5).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Explain the determine blue
= 1/27(9k? – 6k + 9l + 5).
(5.172)
..
The homogeneous equation
(E1, E2 + 2)z½(k, l) = 0
(5.173)
has the solution
Zn (k, l) = (-2)*f(e – k).
(5.174)
Therefore, the general solution of equation (5.170) is
Zh (k, l) = zh(k, l) + zp(k, l)
(5.175)
= (-2)* ƒ(l – k) + /27(9k² – 6k + 9l +5).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc0e39b00-7c5c-4b9e-ab5d-9170cd8d91e5%2F0825aea7-3626-4bf8-acdf-667283a5ce5c%2Fdluky4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5.5.2 Example B
Consider the equation
z(k +1, l + 1) + 2z(k, l) = k² + l +1,
(5.170)
or
(E1 E2 + 2)z(k, l) = k² + l + 1.
(5.171)
The particular solution is
Zp(k, l) =
1
(k² + l +1)
E, Eg + 2
E1E2 + 2
1
1
- (k² + l + 1)
31+ 1/3(A1 + A2 + A¡A2)
= 1/3[1–1/3(A1 + A2 + A¡A2)
+ 1/9(A} + A3 + AA} + 2A¡A2 + 2A{A2
+2A,A3)+ · · ·](k² +l+1)
= 1/27(9k? – 6k + 9l + 5).
(5.172)
..
The homogeneous equation
(E1, E2 + 2)z½(k, l) = 0
(5.173)
has the solution
Zn (k, l) = (-2)*f(e – k).
(5.174)
Therefore, the general solution of equation (5.170) is
Zh (k, l) = zh(k, l) + zp(k, l)
(5.175)
= (-2)* ƒ(l – k) + /27(9k² – 6k + 9l +5).
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