What is the difference between theory 7 and theory 10 Give me a reason for that? Eq1 in the second picture Theorem 7.If k,1 are even and o is odd positive integersand (A+B+ C) +1#D, then Eq.(1) has no prime periodtwo solution.Proof.Following the proof of Theorem 5, we deduce thatif k, 1 are even and o is odd positive integers, then x, =Xn-k = Xn-1 and X+1 = Xn-o. It follows from Eq.(1) thatbP= (A+B+ C) Q+ DP –(16)(e – d)'andbQ= (A+B+C) P+DQ –(17)(e – d)'By subtracting (17) from (16), we get(P- Q) [(A+B+ C+1) – D]= 0,Since (A+B+C+1) – D #0, then P= Q. This is acontradiction. Thus, the proof is now completed.OThus, we deduce that(P+ Q)² 4PQ.(28)Substituting (25) and (26) into (28), we get the condition(20). Thus, the proof is now completed.OTheorem 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) 0.valid,0,e B (A+C)(e+d) [K2 +(A+C)] [d K2 – e (A+C)]² 'PQ=(33)(1- (B+D)), provided (B+ D)

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What is the difference between theory 7 and theory 10 Give me a reason for that? Eq1 in the second picture Theorem 7.If k,1 are even and o is odd positive integersand (A+B+ C) +1#D, then Eq.(1) has no prime periodtwo solution.Proof.Following the proof of Theorem 5, we deduce thatif k, 1 are even and o is odd positive integers, then x, =Xn-k = Xn-1 and X+1 = Xn-o. It follows from Eq.(1) thatbP= (A+B+ C) Q+ DP –(16)(e – d)'andbQ= (A+B+C) P+DQ –(17)(e – d)'By subtracting (17) from (16), we get(P- Q) [(A+B+ C+1) – D]= 0,Since (A+B+C+1) – D #0, then P= Q. This is acontradiction. Thus, the proof is now completed.OThus, we deduce that(P+ Q)² > 4PQ.(28)Substituting (25) and (26) into (28), we get the condition(20). Thus, the proof is now completed.OTheorem 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)isd (1– (B+D)) –(B+D)provided(A+C) > 0.valid,< 1and- eProof.If 1 is even and k, o are odd positive integers, thenIt follows from Eq.(1)Xn = Xn-1andXn+1 = Xn-k= Xn-o.thatbPP= (A+C) Q+ (B+D) P –(30)(e Q- dP)'andbQQ= (A+C)P+(B+D) Q(31)(e P- dQ)'Consequently, we getbP+ Q=(32)[d (1 – (B+D)) – e (A+C)]'where d (1- (B+D)) – e (A+C) >0,e B (A+C)(e+d) [K2 +(A+C)] [d K2 – e (A+C)]² 'PQ=(33)(1- (B+D)), provided (B+ D) < 1.where K2Substituting (32) and (33) into (28), we get the condition(29). Thus, the proof is now completed.O The objective of this article is to investigate somequalitative behavior of the solutions of the nonlineardifference equationbxn-k[dxp-k– exn=1](1)Xn+1Axn+ Bxn–k+ Cxr-1+DXn-ot= U:0,1,2,....where the coefficients A, B, C, D, b, d, e E (0,∞), whilek, 1 and o are positive integers. The initial conditionsX-0,…., 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,•..)

Accepted Answer

Given nonlinear difference equation is
xn+1=Axn+Bn-k+Cxn-l+Dxn-σ+bxn-kdxn-k-exn-l               (1)…