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application of this method considers the Clairaut difference equation As an Yk = kAyk + f(Ayk), (6.52)

application of this method considers the Clairaut difference equation As an Yk = kAyk + f(Ayk), (6.52)

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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ull stc ksa 3:03 PM C @ 1 40% As an application of this method considers the Clairaut difference equation Yk = kAyk + f (Ayk), (6.52) Cancel Actual Size (434 KB) Choose
Transcribed Image Text:ull stc ksa 3:03 PM C @ 1 40% As an application of this method considers the Clairaut difference equation Yk = kAyk + f (Ayk), (6.52) Cancel Actual Size (434 KB) Choose
6.4.1 Example Let the function f be equal to (Ayk)²; therefore, equation (6.52) becomes Yk = kxk + x, (6.61) where we have substituted x = Ayk. Operating with A gives (k +1)A¤k + 2xrAær + (Aæx)² = = 0. (6.62) Thus, we conclude that either Axk = 0 and Yk = ck + c², (6.63) %3D or Ark + 2xk + k +1= xk+1 + xk + k +1 = 0. (6.64) The solution to the last equation is = c(-1)* – 1/½k – 1/4, (6.65) which gives for equation (6.61) the second solution Yk = [c(-1)* – 1/4]² – 1¼k². (6.66)
Transcribed Image Text:6.4.1 Example Let the function f be equal to (Ayk)²; therefore, equation (6.52) becomes Yk = kxk + x, (6.61) where we have substituted x = Ayk. Operating with A gives (k +1)A¤k + 2xrAær + (Aæx)² = = 0. (6.62) Thus, we conclude that either Axk = 0 and Yk = ck + c², (6.63) %3D or Ark + 2xk + k +1= xk+1 + xk + k +1 = 0. (6.64) The solution to the last equation is = c(-1)* – 1/½k – 1/4, (6.65) which gives for equation (6.61) the second solution Yk = [c(-1)* – 1/4]² – 1¼k². (6.66)
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