6.5.9 Example I Let N be a positive number and consider the following nonlinear difference equation: Yk+1 = /½(Yk – N/Yk). (6.120) Using the fact that cot2 0 (cot (20) = - 1 (6.121) %3D 2 cot 0 we can make the substitution yk VN cot xk, reducing equation (6.120) to the form cot xk+1 = cot (2ak). (6.122) A solution of this latter equation is Xk+1 2xk. (6.123) Therefore, Xk = c2k (6.124) and VN cot(c2*).) (6.125) Yk =
6.5.9 Example I Let N be a positive number and consider the following nonlinear difference equation: Yk+1 = /½(Yk – N/Yk). (6.120) Using the fact that cot2 0 (cot (20) = - 1 (6.121) %3D 2 cot 0 we can make the substitution yk VN cot xk, reducing equation (6.120) to the form cot xk+1 = cot (2ak). (6.122) A solution of this latter equation is Xk+1 2xk. (6.123) Therefore, Xk = c2k (6.124) and VN cot(c2*).) (6.125) Yk =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Transcribed Image Text:6.5.9 Example I
Let N be a positive number and consider the following nonlinear difference
equation:
Yk+1 = 1/2(Yk – N/Yk).
(6.120)
Using the fact that
cot2 0 – 1
(cot (20)
(6.121)
2 cot 0
we can make the substitution yk =
VN cot x, reducing equation (6.120) to
the form
cot 2k41 = cot(2aR)
(6.122)
A solution of this latter equation is
Xk+1 =
2xk.
(6.123)
Therefore,
c2k
(6.124)
and
Yk = VN cot(c2*).
)
(6.125)
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