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P= AQ+(B+C+D) P – (14) (e – d) - nd b Q= AP+ (B+C+D) Q – (15) - (e – d) By subtracting (15) from (14), we get (P- Q) [(A+1) – (B+C+D) ] =0. |

P= AQ+(B+C+D) P – (14) (e – d) - nd b Q= AP+ (B+C+D) Q – (15) - (e – d) By subtracting (15) from (14), we get (P- Q) [(A+1) – (B+C+D) ] =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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Theorem 6.If k,1 and o are both odd positive integers and A+17 B+C+D,then Eq. (1) has no prime period two solution. Proof.Following the proof of Theorem 5, we deduce that if k, 1 and o are both odd positive integers, then xn+1 = 6. Xn-k= Xn–1= Xn-o• It follows from Eq.(1) that b P= AQ+ (B+C+D) P – (14) (е — d)' and b Q= AP+(B+C+D) Q – (e (15) - d) By subtracting (15) from (14), we get P- Q) [(A+1) – (B+C+D)]=0. Since (A+1) – contradiction. Thus, the proof is now completed.O (B+C+D) +0, then P= Q. This is a
Transcribed Image Text:Theorem 6.If k,1 and o are both odd positive integers and A+17 B+C+D,then Eq. (1) has no prime period two solution. Proof.Following the proof of Theorem 5, we deduce that if k, 1 and o are both odd positive integers, then xn+1 = 6. Xn-k= Xn–1= Xn-o• It follows from Eq.(1) that b P= AQ+ (B+C+D) P – (14) (е — d)' and b Q= AP+(B+C+D) Q – (e (15) - d) By subtracting (15) from (14), we get P- Q) [(A+1) – (B+C+D)]=0. Since (A+1) – contradiction. Thus, the proof is now completed.O (B+C+D) +0, then P= Q. This is a
Theorem 5.If k,1 and o are both even positive integers, then Eq. (1) has no prime period two solution. Proof.Assume that there exists distinct positive solutions P,Q, P, Q,.. of prime period two of Eq.(1). If k,1 and o are both even positive integers, then x, = Xp-k= Xn-1= Xn-o. from Eq. (1) that It follows P=(A+B+C+D) Q – (e d) (12) | and b Q= (A+B+C+D)P – (13) (e – d)' | By subtracting (13) from (12), we get (P- Q) [A+B+C+D+1]=0. || Since A+ B+C +D+1 0, then P: contradiction. Thus, the proof is now completed.O Q. This is a 6. The objective of this article is to investigate some qualitative behavior of the solutions of the nonlinear difference equation bxr- k [dxn-k- exn-1] (1) Xn+1 Axn+ Bxn–k+ Cxn–1+Dxn-o + n = 0,1, 2, .... where the coefficients A, B, C, D, b, d, e € (0,0), while k,1 and o are positive integers. The initial conditions X-,..., X_1,..., X_k ….., X_1, Xo are arbitrary positive real numbers such that k <1< 0. Note that the special cases of Eq. (1) have been studied in [1] when B=C= D=0, and k= 0,1= 1, b is replaced by – b and in [27] when B=C= D=0, and k= 0, b is replaced by – b and in [33] when B = C = D = 0, 1= 0 and in [32] when A=C= D=0, 1=0, b is replaced by – b. 6.
Transcribed Image Text:Theorem 5.If k,1 and o are both even positive integers, then Eq. (1) has no prime period two solution. Proof.Assume that there exists distinct positive solutions P,Q, P, Q,.. of prime period two of Eq.(1). If k,1 and o are both even positive integers, then x, = Xp-k= Xn-1= Xn-o. from Eq. (1) that It follows P=(A+B+C+D) Q – (e d) (12) | and b Q= (A+B+C+D)P – (13) (e – d)' | By subtracting (13) from (12), we get (P- Q) [A+B+C+D+1]=0. || Since A+ B+C +D+1 0, then P: contradiction. Thus, the proof is now completed.O Q. This is a 6. The objective of this article is to investigate some qualitative behavior of the solutions of the nonlinear difference equation bxr- k [dxn-k- exn-1] (1) Xn+1 Axn+ Bxn–k+ Cxn–1+Dxn-o + n = 0,1, 2, .... where the coefficients A, B, C, D, b, d, e € (0,0), while k,1 and o are positive integers. The initial conditions X-,..., X_1,..., X_k ….., X_1, Xo are arbitrary positive real numbers such that k <1< 0. Note that the special cases of Eq. (1) have been studied in [1] when B=C= D=0, and k= 0,1= 1, b is replaced by – b and in [27] when B=C= D=0, and k= 0, b is replaced by – b and in [33] when B = C = D = 0, 1= 0 and in [32] when A=C= D=0, 1=0, b is replaced by – b. 6.
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