Example D The system of equations (E – 1)uk + 2Evk = 0, -2uk + (E – 1)v = a*, (4.349) when solved for vk, gives the equation (E+1)²vµ = (a – 1)a*, (4.350) the solution of which is a – 1 Uk = (c1 + c2k)(–1)* + (4.351) (a + 1)2 a. Now, from the second of equations (4.349), we have Uk = ½(E – 1)vk – 12a*. (4.352) - The substitution of vk from equation (4.351) into the right-hand side of tion (4.351) gives equa- 2ak+1 Uk = (-/½)[(2c1 + c2)+ 2c2k](-1)k (4.353) - (a + 1)2 '
Example D The system of equations (E – 1)uk + 2Evk = 0, -2uk + (E – 1)v = a*, (4.349) when solved for vk, gives the equation (E+1)²vµ = (a – 1)a*, (4.350) the solution of which is a – 1 Uk = (c1 + c2k)(–1)* + (4.351) (a + 1)2 a. Now, from the second of equations (4.349), we have Uk = ½(E – 1)vk – 12a*. (4.352) - The substitution of vk from equation (4.351) into the right-hand side of tion (4.351) gives equa- 2ak+1 Uk = (-/½)[(2c1 + c2)+ 2c2k](-1)k (4.353) - (a + 1)2 '
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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Topic Video
Question
Explain the determine
*
(4.353)
(a + 1)2"](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8996ece0-6cdf-4a3a-b245-e8d4715722e7%2Fe6f37d29-6465-4622-9c8c-231fe565b30e%2Fhsdr33_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Difference Equations
Example D
The system of equations
(E – 1)uk + 2Evk = 0,
— 2ик + (Е — 1)uz — а*,
(4.349)
= a.
when solved for vk, gives the equation
(E + 1)²v% = (a – 1)a*,
(4.350)
the solution of which is
а — I
: (c1 + c2k)(-1)k
(4.351)
Uk =
(а +
Now, from the second of equations (4.349), we have
Uk =1/2(E – 1)VK
/2a*.
(4.352)
The substitution of vk from equation (4.351) into the right-hand side of
tion (4.351) gives
equa-
2ak+1
uk = (-1/2)[(2c1 + c2) + 2c2k](-1)*
(4.353)
(a + 1)2"
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