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The main aim of this study is to exhibit some cases on the periodic character of the positive solutions of the rational difference equation aSn-g + bSn-r + cSn-s d.Sn-q+ eSn-r + fS. Sn+1 Sn-p (1) n-s b-u where a, b, c, d, e, f € (0, 0). The initial conditions S-p, S-p+1,...,S=q, S-q+1,...,S=r, S-r+1,...,S-s,...,S-s+1,...,S-1 and So are arbitrary positive real numbers such that p > q > r > s > 0.

The main aim of this study is to exhibit some cases on the periodic character of the positive solutions of the rational difference equation aSn-g + bSn-r + cSn-s d.Sn-q+ eSn-r + fS. Sn+1 Sn-p (1) n-s b-u where a, b, c, d, e, f € (0, 0). The initial conditions S-p, S-p+1,...,S=q, S-q+1,...,S=r, S-r+1,...,S-s,...,S-s+1,...,S-1 and So are arbitrary positive real numbers such that p > q > r > s > 0.

Advanced Engineering Mathematics
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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The main aim of this study is to exhibit some cases on the periodic character of the positive solutions of the rational difference equation aSn-q + bSn-r + cSn-s dSn Sn+1 = Sn-p (1) + eSn-r + fSn-s ) ' -q where a, b, c, d, e, ƒ€ (0, 0). The initial conditions S-p, S-p+1;.-,S-q, S-q+1;..,S-r, S-r+1,...,S-s,...,S_s+1,...,S_1 and So are arbitrary positive real numbers such that p > q > r > s > 0. е,
Transcribed Image Text:The main aim of this study is to exhibit some cases on the periodic character of the positive solutions of the rational difference equation aSn-q + bSn-r + cSn-s dSn Sn+1 = Sn-p (1) + eSn-r + fSn-s ) ' -q where a, b, c, d, e, ƒ€ (0, 0). The initial conditions S-p, S-p+1;.-,S-q, S-q+1;..,S-r, S-r+1,...,S-s,...,S_s+1,...,S_1 and So are arbitrary positive real numbers such that p > q > r > s > 0. е,
(d+e+ f)¢b = (a + b+c)ø?. Case 2. Suppose the positive integers p,q,r and s are even. In this case Sn = Sn-p = Sn-q = Sn-r = Sn-8. From Equation (1), we have Ø = v ( @+b+ c)½ (d+e+ f)b, (a +b+ c)¢ (d+e+ f)¢, Thus, (d+ e+ f)øv½ = (a + b+ c)µ², (4) and (5) By subtracting (4) from (5), we deduce that (a + b+ c)(ø² – b?) = 0. Hence, we have (a + b+ c)(6? – ?) = 0. If ø + y, then ((a + b+ c) = 0. Since a, b andc are nonzero positive real numbers, thus (a+b+c) # 0. This implies O = v. This contradicts the hypothesis o + v.
Transcribed Image Text:(d+e+ f)¢b = (a + b+c)ø?. Case 2. Suppose the positive integers p,q,r and s are even. In this case Sn = Sn-p = Sn-q = Sn-r = Sn-8. From Equation (1), we have Ø = v ( @+b+ c)½ (d+e+ f)b, (a +b+ c)¢ (d+e+ f)¢, Thus, (d+ e+ f)øv½ = (a + b+ c)µ², (4) and (5) By subtracting (4) from (5), we deduce that (a + b+ c)(ø² – b?) = 0. Hence, we have (a + b+ c)(6? – ?) = 0. If ø + y, then ((a + b+ c) = 0. Since a, b andc are nonzero positive real numbers, thus (a+b+c) # 0. This implies O = v. This contradicts the hypothesis o + v.
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