Explain the determine green and the inf is here Theorem 12.If k,o are even and 1 is odd positive integers,then Eq. (1) has prime period two solution if the condition(3e- d) (1– C)< (e+d)(A+B+D),(39)is valid, provided C<1 and e (1- C)-d(A+B+D) >0.Proof.If k,o are even and 1 is odd positive integers, thenXn-1. It follows from Eq. (1)Xn = Xn-k = Xn-o and Xn+1thatbQP= (A+B+D) Q+CP(40)(e P- dQ)’andbPQ= (A+B+D) P+CQ –(41)(e Q- dP)Consequently, we getbP+ Q=(42)[e (1-C)-d (A+B+D)]´where e (1 – C) – d (A+B+D) > 0,eb (1–C)PQ=(43)(e+ d) [(1– C) + K4] [e (1 – C )– d K4]²(A+B+D), provided C < 1. Substitutingwhere K4(42) and (43) into (28), we get the condition (39). Thus,the proof is now completed.O The objective of this article is to investigate somequalitative behavior of the solutions of the nonlineardifference equationbxn– kX+1 = Axn+ Bxp–k+CXp-1+Dxp-o+[dxn-k– ex-1(1)n= 0, 1,2, .....where the coefficients A, B, C, D, b, d, e e (0,00), whilek, 1 and o are positive integers. The initial conditionsX-g,..., X_1,..., X_ k, ..., X_1, Xo are arbitrary positive realnumbers such that k < 1 < 0. Note that the special casesof Eq. (1) have been studied in [1] when B= C= D= 0,and k = 0,1= 1, b is replaced byB=C= D=0, and k= 0, b is replaced by – b and in[33] when B = C = D = 0, 1= 0 and in [32] whenA= C= D=0, 1=0, b is replaced by – b.••..- b and in [27] when6.

Name: Explain the determine green and the inf is here Theorem 12.If k,o are
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Description: Explain the determine green and the inf is here Theorem 12.If k,o are even and 1 is odd positive integers,then Eq. (1) has prime period two solution if the condition(3e- d) (1– C) 0.Proo

Given P=A+B+CQ+CP-bQeP-dQ (1) Q=A+B+DP+CQ-bPeQ-dP (2) We…

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Theorem 12.If k, o are even and 1 is odd positive integers, then Eq. (1) has prime period two solution if the condition (3e-d) (1-C) < (e+d) (A+B+D), (39) is valid, provided C < 1 and e (1− C) −d (A+B+D) > 0. Proof. If k,o are even and I is odd positive integers, then Xn = Xn-k = Xn-o and Xn+1 and Xn+1 = Xn-1. It follows from Eq. (1) that and P= (A+B+ D) Q+ CP Q= (A+B+D) P+CQ Consequently, we get - b CP+Q=(1-0)- #1 bQ (e P- dQ)' bP (e Q- dp)* [e (1 − C) −d (A+B+D)]' (40) (41) (42)