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Theorem 10.If 1 is even and k, o are odd positive integers, then Eq. (1) has prime period two solution if the condition (A+C) (3e– d) < (e+d)(1– (B+D)), (29) is valid, (B+D) < 1 provided d (1 – (B+ D)) – e (A+ C) > 0. and

Theorem 10.If 1 is even and k, o are odd positive integers, then Eq. (1) has prime period two solution if the condition (A+C) (3e– d) < (e+d)(1– (B+D)), (29) is valid, (B+D) < 1 provided d (1 – (B+ D)) – e (A+ C) > 0. and

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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How to deduce this equation from Equation 1 Explain to me the method. Show me the steps of determine red and equation 1 is second pic

Theorem 10.If 1 is even and k, o are odd positive integers, then Eq. (1) has prime period two solution if the condition (A+C) (3e- d) < (e+d)(1 – (B+ D)), (29) is valid, provided (B+D) 1 and d (1 – (B+ D)) – e (A+ C) > 0. - Proof.If 1 is even and k, o are odd positive integers, then Xn = Xp-1 and Xn+1 = Xn-k = Xn-o. It follows from Eq.(1) that bP P= (A+C) Q+(B+D) P – (30) (e Q- dP) and bQ Q= (A+C)P+(B+D) Q – (31) (e P- dQ)' Consequently, we get b P+Q= (32) [d (1– (B+D)) – e (A+C)]' where d (1 – (B+D)) – e (A+C) > 0, e b (A+ C) PQ= (e+d)[K2+(A+C)][d K2 – e (A+C)]² (33) (1– (B+ D)), provided (B+ D) < 1. where K2 Substituting (32) and (33) into (28), we get the condition (29). Thus, the proof is now completed.O
Transcribed Image Text:Theorem 10.If 1 is even and k, o are odd positive integers, then Eq. (1) has prime period two solution if the condition (A+C) (3e- d) < (e+d)(1 – (B+ D)), (29) is valid, provided (B+D) 1 and d (1 – (B+ D)) – e (A+ C) > 0. - Proof.If 1 is even and k, o are odd positive integers, then Xn = Xp-1 and Xn+1 = Xn-k = Xn-o. It follows from Eq.(1) that bP P= (A+C) Q+(B+D) P – (30) (e Q- dP) and bQ Q= (A+C)P+(B+D) Q – (31) (e P- dQ)' Consequently, we get b P+Q= (32) [d (1– (B+D)) – e (A+C)]' where d (1 – (B+D)) – e (A+C) > 0, e b (A+ C) PQ= (e+d)[K2+(A+C)][d K2 – e (A+C)]² (33) (1– (B+ D)), provided (B+ D) < 1. where K2 Substituting (32) and (33) into (28), we get the condition (29). Thus, the proof is now completed.O
Thus, we deduce that (P+ Q)² > 4PQ. (28) The objective of this article is to investigate some qualitative behavior of the solutions of the nonlinear difference equation bxn-k [dxn-k - exp-1) Xn+1 = Axn+ Bxn-k+Cxn–1+Dxn-o + n = 0,1,2,.. where the coefficients A, B, C, D, b, d, e e (0,00), while k, 1 and o are positive integers. The initial conditions X-6…, X_1,..., X_k, ..., X_1, Xo are arbitrary positive real numbers such that k <1< o. 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 [33] when B = C = D = 0, 1 = 0 and in [32] when A = C= D=0, 1= 0, b is replaced by – b. (1) b and in %3D
Transcribed Image Text:Thus, we deduce that (P+ Q)² > 4PQ. (28) The objective of this article is to investigate some qualitative behavior of the solutions of the nonlinear difference equation bxn-k [dxn-k - exp-1) Xn+1 = Axn+ Bxn-k+Cxn–1+Dxn-o + n = 0,1,2,.. where the coefficients A, B, C, D, b, d, e e (0,00), while k, 1 and o are positive integers. The initial conditions X-6…, X_1,..., X_k, ..., X_1, Xo are arbitrary positive real numbers such that k <1< o. 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 [33] when B = C = D = 0, 1 = 0 and in [32] when A = C= D=0, 1= 0, b is replaced by – b. (1) b and in %3D
Expert Solution
Step 1

The equation given to us is as follows:

xn+1=Axn+Bxn-k+Cxn-l+Dxn-σ+bxn-kdxn-k-exn-l...................................(1)

We are given that if l is even integer and k, σ are odd integers, then xn=xn-l and xn+1=xn-k=xn-σ. 

Substitute Q=xn=xn-l and P=xn+1=xn-k=xn-σ in equation (1).

P=AQ+BP+CQ+DP+bPdP-eQ

Simplify the RHS.

P=A+CQ+B+DP-bPeQ-dP.............................................(2)

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