6.5.1 Example A Consider the nonlinear equation Yk+2 = (yk+1)²/Yk- (6.74) If we set xk = log yk, then ak satisfies the equation %3D Xk+2 – 2xk+1 + xk = 0, (6.75) NONLINEAR DIFFERENCE EQUATIONS 203 the solution of which is Xk = C1 + c2k (6.76) Therefore, a general solution of equation (6.74) is Yk = ec1+c2k (6.77) or Yk Czec2k C3 > 0. (6.78)
6.5.1 Example A Consider the nonlinear equation Yk+2 = (yk+1)²/Yk- (6.74) If we set xk = log yk, then ak satisfies the equation %3D Xk+2 – 2xk+1 + xk = 0, (6.75) NONLINEAR DIFFERENCE EQUATIONS 203 the solution of which is Xk = C1 + c2k (6.76) Therefore, a general solution of equation (6.74) is Yk = ec1+c2k (6.77) or Yk Czec2k C3 > 0. (6.78)
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:6.5.1
Example A
Consider the nonlinear equation
Yk+2 = (Yk+1)²/yk.
(6.74)
If we set Xk
log yik, then xj, satisfies the equation
Xk+2
2xk+1 + Xk = 0,
(6.75)
NONLINEAR DIFFERENCE EQUATIONS
203
the solution of which is
Xk = C1 + c2k
(6.76)
Therefore, a general solution of equation (6.74) is
Yk = e©i+c2k
(6.77)
or
Czec2k
Сз > 0.
(6.78)
Yk
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