Show me the steps of determine green and inf is here Theorem 14.If 0< A+B+ C+D<1 and e+ d, thenthe equilibrium point x given by (7) of Eq.(1) is globalattractor.Proof.We consider the following functionbyF(x,y, z, w) = Ax+By+Cz+ Dw+(dy– ez)(50)where dy + ez, provided that B(dy – ez)²>bez andC(dy– ez)? + bey>0. It is easy to verify the condition (i)of Theorem 3. Let us now verify the condition (ii) ofTheorem 3 as follows:[F(x,x, x, x) – x] (x- x) =K5 x-|е —dt [Ks – 1] (e- d)Sx(e-d) [K5 – 1]-b{xce-d1(51)[K5 – 1]'where K5 = (A+B+C+D). Since 0 < K5 < 1 and e+ d,then we deduce from (51) that[F(x, х, х, х) — х] (х— х) <0.(52)According to Theorem 3, is global attractor. Thus, theproof is now completed.OOn combining the two Theorems 4 and 14, we have theresult. The objective of this article is to investigate somequalitative behavior of the solutions of the nonlineardifference equationbxn-k[dxp-k- exn-i](1)Xn+1Axn+ Bxn-k+ Cxp–1+DXn-o +n = 0,1,2, ..where the coefficients A, B, C, D, b, d, e E (0,0), whilek, 1 and o are positive integers. The initial conditionsX-,..., 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 by – b and in [27] whenB=C= D=0, and k= 0,b is replaced by – b and in[33] when B = C = D = 0, 1 :A=C=D=0, 1=0, b is replaced by – b.0 and in [32] when-bX=[(A+B+C+D) – 1](e – d)(7)LetusnowintroducecontinuousfunctionF: (0, 00)* → (0,0) which is defined byaibunTheorem 3.[12]. Consider the difference equation (2).Let x e I be an equilibrium point of Eq. (2). Suppose alsothat(i) Farguments.is a nondecreasing function in each of its(ii) The function F satisfies the negative feedbackproperty[F(x, x, x, x) – x] (x– x) < 0 for all xE I- {x},where I is an open interval of real numbers. Then x isglobal attractor for all solutions of Eq. (2).

Given : Equilibrium point x~=bA+B+C+D-1e-d 7 To verify : condition (ii) of…

Theorem 14.If 0< A+B+C+D<1 and e + d, then the equilibrium point x given by (7) of Eq.(1) is global attractor. Proof. We consider the following function by (50) (dy – ez)' F(x,y, z, w) = Ax+By+Cz+ Dw+ where dy + ez, provided that B(dy – ez)²>bez and C(dy– ez)2 + bey>0. It is easy to verify the condition (i) of Theorem 3. Let us now verify the condition (ii) of Theorem 3 as follows: b [F(x,x, x, x) – x] (x– F) = {. Kgx- -d е — b {x- [Ks – 1] (e- d) X Sx(e-d)[Kg – 1] - b e-d 1 (51) [K5 – 1]' where K5 = (A+B+C+D). Since 0 < K5 < 1 and e# d, then we deduce from (51) that [F(х, х, х, х) — х] (х— X) <0. (52) According to Theorem 3, is global attractor. Thus, the proof is now completed.O On combining the two Theorems 4 and 14, we have the result.