6.5.4 Example D Consider the equation VYk+1 = k/Yk. (6.87) Letting ak = V Yk gives Xk+1 = kxk, (6.88) the solution of which is c(k – 1)!. (6.89) Xk = Therefore, Yk = [c(k – 1)!]². (6.90) We can obtain this solution to equation (6.87) by a second method. Taking the logarithm of both sides of equation (6.87) gives Zk+1 - Zk = log k, (6.91) where zk = 1/2(log yk). This last equation is of the form Zk+1 - R (6.92) and can be solved by using the techniques of Chapter 1. Its solution is Zk = log[c(k – 1)!]. (6.93) %3D and, consequently, the result given by equation (6.90) is obtained.
6.5.4 Example D Consider the equation VYk+1 = k/Yk. (6.87) Letting ak = V Yk gives Xk+1 = kxk, (6.88) the solution of which is c(k – 1)!. (6.89) Xk = Therefore, Yk = [c(k – 1)!]². (6.90) We can obtain this solution to equation (6.87) by a second method. Taking the logarithm of both sides of equation (6.87) gives Zk+1 - Zk = log k, (6.91) where zk = 1/2(log yk). This last equation is of the form Zk+1 - R (6.92) and can be solved by using the techniques of Chapter 1. Its solution is Zk = log[c(k – 1)!]. (6.93) %3D and, consequently, the result given by equation (6.90) is obtained.
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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Explain the determine blue
Transcribed Image Text:6.5.4 Example D
Consider the equation
Yk+1 = k/Yk•
(6.87)
Letting xk =
VYk gives
Xk+1 =
kak,
(6.88)
the solution of which is
c(k – 1)!.
(6.89)
Therefore,
Yk = [c(k – 1)!?.
(6.90)
We can obtain this solution to equation (6.87) by a second method. Taking
the logarithm of both sides of equation (6.87) gives
Zk+1 – 2k = log k,
(6.91)
where z = 1/2(log yk). This last equation is of the form
Zk+1 – PrZk = Rµ
(6.92)
and can be solved by using the techniques of Chapter 1. Its solution is
Zk = log[c(k – 1)
(6.93)
and, consequently, the result given by equation (6.90) is obtained.
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