# Analytic Solutions and Stability of Sixth Order Difference Equations

Authors: H. S. Alayachi; M. S. M. Noorani; A. Q. Khan; M. B. Almatrafi
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

In the present paper, the global attractor, local stability, and boundedness of the solution of sixth order difference equations are investigated analytically and numerically. The exact solutions of three equations are presented by utilizing Fibonacci sequence. We also analyse the periodicity of a sixth order difference equation. The considered difference equations are given byyn+1=Ayn−1±Byn−1yn−3/Cyn−3±Dyn−5,n=0,1,…, where the initial conditions y−5,y−4,y−3,y−2,y−1, and y0 are arbitrary real numbers and the values A,B,C, and D are defined as positive real numbers.


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## Body

## 1. Introduction

In a contemporary study, the theory of difference equations has been studied by a huge number of researchers. This can be attributed to the importance of this field in modelling a large number of natural phenomena. Difference equations are used in modelling some real-life problems appeared in biology, physics, economy, engineering, etc. Difference equations become apparent in the study of discretization methods for differential equations. Some results in the theory of difference equations have been obtained in the corresponding results of differential equations as more or less natural discrete analogues. Some recent studies of the dynamic of difference equations are given as follows. Almatrafi [2] obtained the exact solutions of the following systems of the difference equations:(1)xn+1=xn−1yn−3yn−1−1−xn−1yn−3,yn+1=yn−1xn−3xn−1±1±yn−1xn−3.Alotaibi [5] analysed the global stability and examined the periodic solution of the following difference equation:(2)yn+1=Aym+α1ym−1+α2ym−2+α3ym−3+α4ym−4+α5ym−5β1ym−1+β2ym−2+β3ym−3+β4ym−4+β5ym−5.Also, in [8], the authors dealt with the difference equation and considered some special cases of(3)xn+1=αxn+βxnxn−1+γxn−1xnAxn+Bxnxn−1+Cxn−1.Cinar [10] investigated the solution of the difference equation:(4)xn+1=axn−11+bxnxn−1.El-Dessoky [11] studied the qualitative properties of the fifth order nonlinear difference equation:(5)yn+1=αyn+βynyn−1Ayn−4+Byn−3.Bektešević et al. [13] gave a description about the global stability of three special cases of the following recursive equation:(6)xn+1=αxn2+βxnxn−1+γxn−1Axn2+Bxnxn−1+Cxn−1.Ibrahim [14] offered some relevant results of the difference equation:(7)xn+1=xnxn−2xn−1a+bxnxn−2.The authors in [18] presented some theoretical investigations for some of the formulae of(8)xn+1=axn−1b+cxn−1yn,yn+1=αyn−1β+γyn−1xn.One can see the works of [1, 3, 4, 6] and [7, 9, 12, 17, 19, 20, 21, 22, 23, 24] to obtain more information about rational difference equations and systems.The prime objective of this study is to explore the qualitative behavior of the solutions of the following recursive equation:(9)yn+1=Ayn−1+Byn−1yn−3Cyn−3+Dyn−5,n=0,1,…,where the initial conditions y−5,y−4,y−3,y−2,y−1, and y0 are arbitrary nonzero real numbers. The constants A,B,C, and D are assumed to be positive.


## 2. Local Stability of the Equilibrium Point of Equation (9)

This section focuses on discovering the local stability of the solutions around the equilibrium point of equation (9). Equation (9) has a unique equilibrium point given by(10)y¯=Ay¯+By¯2Cy¯+Dy¯.IfC+D1−A≠B, then the only positive equilibrium point of equation (9) is given by(11)y¯=0.Letf:0,∞3⟶0,∞ be a continuous function defined by(12)fu,v,w=Au+BuvCv+Dw.Hence, we can obtain the following partial derivatives:(13)∂fu,v,w∂u=A+BvCv+Dw,∂fu,v,w∂v=BDuwCv+Dw2,∂fu,v,w∂w=−BDuvCv+Dw2.Evaluating the previous derivatives aty¯ gives(14)∂fy¯,y¯,y¯∂u=A+BC+D,∂fy¯,y¯,y¯∂v=BDC+D2,∂fy¯,y¯,y¯∂w=−BDC+D2.The linearized equation of equation (9) about y¯ is now expressed by(15)yn+1−A+BC+Dyn−1−BDC+D2yn−3+BDC+D2=0.Theorem 1.
Assume that(16)BC+3D<1−AC+D2.Then, the positive fixed point of equation (9) is locally asymptotically stable.Proof.
Theorem A in [15] ensures that equation (9) is asymptotically stable if(17)A+BC+D+BDC+D2+BDC+D2<1,that is,(18)A+BC+D+2BDC+D2<1,or,(19)BC+3DC+D2<1−A,which is the required condition.


## 3. Global Attractivity of the Equilibrium Point of Equation (9)

This section is devoted to investigate the global attractivity of the solutions of equation (9).Theorem 2.
The fixed pointy¯ of equation (9) is a global attractor if C1−A≠B.Proof.
Letα and β be a real numbers and assume that f:α,β3⟶α,β be a function defined by equation (12). Then, we can easily observe that the function fu,v,w is increasing in u and v and is decreasing in w. Suppose that m,M is a solution of the system(20)M=fM,M,m,m=fm,m,M.
Then, from equation (9), one can notice that(21)M=AM+BM2CM+Dm,m=Am+Bm2Cm+DM,which leads to(22)CM2+DmM=ACM2+ADmM+BM,(23)Cm2+DmM=ACm2+ADmM+Bm2.
Subtracting equation (23) from (22) gives(24)C1−AM2−m2=BM2−m2,C1−A≠B.
Thus,(25)M=m.
It follows by Theorem B in [16] that y¯ is a global attractor of equation (9). Therefore, the proof is complete.


## 4. Boundedness of Solutions of Equation (9)

The boundedness of solutions of equation (9) is deeply considered in this section.Theorem 3.
Every solution of (9) is bounded if A+B/C<1.Proof.
Letynn=−5∞ be a solution of equation (9). It follows from equation (9) that(26)yn+1=Ayn−1+Byn−1yn−3Cyn−3+Dyn−5≤Ayn−1+Byn−1yn−3Cyn−3=A+BCyn−1.
Consequently,(27)yn+1≤yn−1for all n≥0.


## 5. Special Cases of Equation (9)

Our main target in this section is to find specific forms for the solutions of some special cases of equation (9) when A,B,C, and D are positive integers. We also aim to give some numerical examples confirming the theoretical work.


### 5.1. On the Equationyn+1=yn−1+yn−1yn−3/yn−3+yn−5

In this section, we present the exact solutions of equation (9):(28)yn+1=yn−1+yn−1yn−3yn−3+yn−5,where the initial conditions y−5,y−4,y−3,y−2,y−1, and y0 are arbitrary real numbers.Theorem 4.
Letxnn=−5∞ be a solution to equation (28) satisfying y−5=k, y−4=h, y−3=t, y−2=s, y−1=l, and y0=r. Then, for n=0,1,2,…,(29)y4n−3=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t,y4n−2=s∏i=0n−1F2i+2h+F2i+3sF2i+1r+F2isF2i+1h+F2i+2sF2ir+F2i−1s,y4n−1=l∏i=0n−1F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t,y4n=r∏i=0n−1F2i+2h+F2i+3sF2i+3r+F2i+2sF2i+1h+F2i+2sF2i+2r+F2i+1s,where Fmm=0∞=0,1,1,2.3,5,8,13,…Proof.
Forn=0, the result holds. Now, suppose that n>0 and that our assumption holds for n−1, that is,(30)y4n−9=l∏i=0n−3F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t,y4n−8=r∏i=0n−3F2i+2h+F2i+3sF2i+3r+F2i+2sF2i+1h+F2i+2sF2i+2r+F2i+1s,y4n−7=t∏i=0n−2F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t,y4n−6=s∏i=0n−2F2i+2h+F2i+3sF2i+1r+F2isF2i+1h+F2i+2sF2ir+F2i−1s,y4n−5=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t,y4n−4=r∏i=0n−2F2i+2h+F2i+3sF2i+3r+F2i+2sF2i+1h+F2i+2sF2i+2r+F2i+1s.
Now, it follows from equation (28) that(31)y4n−3=y4n−5+y4n−5y4n−7y4n−7+y4n−9=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t+l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2t/F2i+1k+F2i+2tF2i+2l+F2i+1tt∏i=0n−2F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1tt∏i=0n−2F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1t+l∏i=0n−3F2i+2k+F2i+3tF2i+3l+F2i+2t/F2i+1k+F2i+2tF2i+2l+F2i+1t=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t+l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2t/F2i+1k+F2i+2tF2i+2l+F2i+1tl∏i=0n−2F2i+2k+F2i+3t/F2i+1k+F2i+2tl∏i=0n−2F2i+2k+F2i+3t/F2i+1k+F2i+2t+l∏i=0n−3F2i+2k+F2i+3t/F2i+1k+F2i+2t=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t+l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2n−2k+F2n−1t/F2i+1k+F2i+2tF2i+2l+F2i+1tF2n−3k+F2n−2tF2n−2k+F2n−1t/F2n−3k+F2n−2t+1=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t1+F2n−2k+F2n−1t/F2n−3k+F2n−2tF2n−2k+F2n−1t/F2n−3k+F2n−2t+1=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t1+F2n−2k+F2n−1tF2n−2k+F2n−1t+F2n−3k+F2n−2t=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t1+F2n−2k+F2n−1tF2n−1k+F2nt=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1tF2n−2k+F2n−1t+F2n−1k+F2ntF2n−1k+F2nt=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1tF2nk+F2n+1tF2n−1k+F2nt.
Then,(32)y4n−3=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t.
Also, we get from equation (28) that(33)y4n−1=y4n−3+y4n−3y4n−5y4n−5+y4n−7=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1tl∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2t/F2i+1k+F2i+2tF2i+2l+F2i+1tl∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2t/F2i+1k+F2i+2tF2i+2l+F2i+1t+t∏i=0n−2F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1tl∏i=0n−2F2i+3l+F2i+2t/F2i+2l+F2i+1tl∏i=0n−2F2i+3l+F2i+2t/F2i+2l+F2i+1t+t∏i=0n−2F2i+1l+F2it/F2il+F2i−1t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+lt∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1t∏i=0n−2F2i+2l+F2i+1t/F2i+3l+F2i+2tl∏i=0n−2F2i+3l+F2i+2t/F2i+2l+F2i+1t+t∏i=0n−2F2i+1l+F2it/F2il+F2i−1t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+lt∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1tl+lF2n−2l+F2n−3t/F2n−1l+F2n−2t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1t1+F2n−2l+F2n−3t/F2n−1l+F2n−2t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1t1+F2n−2l+F2n−3t/F2n−1l+F2n−2tF2n−1l+F2n−2tF2n−1l+F2n−2t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1tF2n−1l+F2n−2tF2n−1l+F2n−2t+F2n−2l+F2n−3t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1tF2n−1l+F2n−2tF2nl+F2n−1t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t1+F2n−1l+F2n−2tF2nl+F2n−1t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1tF2nl+F2n−1t+F2n−1l+F2n−2tF2nl+F2n−1t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1tF2n+1l+F2ntF2nl+F2n−1t=l∏i=0n−1F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t.
Thus, other relations can be similarly proved.


### 5.2. On the Equationyn+1=yn−1+yn−1yn−3/yn−3−yn−5

This section deals with the solutions of the following equation:(34)yn+1=yn−1+yn−1yn−3yn−3−yn−5,where the initial conditions y−5,y−4,y−3,y−2,y−1, and y0 are nonzero arbitrary real numbers.Theorem 5.
Letxnn=−5∞ be a solution to equation (34) satisfying y−5=k, y−4=h, y−3=t, y−2=s, y−1=l, and y0=r. Then, for n=0,1,2,…,(35)y4n−3=Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−Fnktt−kl−t,y4n−2=Fnr−Fn−2sFn+1r−Fn−1sFn+1s−Fn−1hFn+2s−Fnhsr−ss−h,y4n−1=Fn+2l−FntFn+1l−Fn−1tFn+2t−FnkFn+1t−Fn−1ktt−kl−t,y4n=Fn+2s−FnhFn+1s−Fn−1hFn+2r−FnsFn+1r−Fn−1ssr−ss−h.Proof.
Forn=0, the result holds. Now, suppose that n>0 and that our assumption holds for n−1, that is,(36)y4n−9=Fnl−Fn−2tFn−1l−Fn−3tFnt−Fn−2kFn−1t−Fn−3ktt−kl−t,y4n−8=Fns−Fn−2hFn−1s−Fn−3hFnr−Fn−2sFn−1r−Fn−3ssr−ss−h,y4n−7=Fn−1l−Fn−3tFnl−Fn−2tFnt−Fn−2kFn+1t−Fn−1ktt−kl−t,y4n−6=Fn−1r−Fn−3sFnr−Fn−2sFns−Fn−2hFn+1s−Fn−1hsr−ss−h,y4n−5=Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2ktt−kl−t,y4n−4=Fn+1s−Fn−1hFns−Fn−2hFn+1r−Fn−1sFnr−Fn−2ssr−ss−h.
Now, it can be seen from equation (34) that(37)y4n−3=y4n−5+y4n−5y4n−7y4n−7−y4n−9=Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2ktt−kl−t+Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2k/tt−kl−tFn−1l−Fn−3tFnl−Fn−2tFnt−Fn−2kFn+1t−Fn−1k/tt−kl−tFn−1l−Fn−3tFnl−Fn−2tFnt−Fn−2kFn+1t−Fn−1k/tt−kl−t−Fnl−Fn−2tFn−1l−Fn−3tFnt−Fn−2kFn−1t−Fn−3k/tt−kl−t=Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2ktt−kl−t+Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2k/tt−kl−tFn−1l−Fn−3tFnl−Fn−2tFnt−Fn−2kFn+1t−Fn−1k/tt−kl−tFn−1l−Fn−3tFnl−Fn−2tFnt−Fn−2k/tt−kl−tFn+1t−Fn−1k−Fn−1t−Fn−3k=Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2ktt−kl−t+Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2kFn+1t−Fn−1ktt−kl−tFn+1t−Fn−1k−Fn−1t−Fn−3k=Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2kFn+1t−Fn−1k2−Fn−1t−Fn−3ktt−kl−tFn+1t−Fn−1k−Fn−1t−Fn−3k=Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−Fnktt−kl−t.
It can also be observed that(38)y4n−1=y4n−3+y4n−3y4n−5y4n−5−y4n−7=Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−Fnktt−kl−t+Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−Fnk/tt−kl−tFn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2k/tt−kl−tFn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2k/tt−kl−tFn−1l−Fn−3tFnl−Fn−2tFnt−Fn−2kFn+1t−Fn−1k/tt−kl−t=Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−Fnktt−kl−t+Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−Fnk/tt−kl−tFn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2k/tt−kl−tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2k/tt−kl−tFn+1l−Fn−1t−Fn−1l−Fn−3t=Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−Fnktt−kl−t+Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−FnkFn+1l−Fn−1ttt−kl−tFn+1l−Fn−1t−Fn−1l−Fn−3t=Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−FnkFn+1l−Fn−1t2−Fn−1l−Fn−3ttt−kl−tFn+1l−Fn−1t−Fn−1l−Fn−3t=Fn+2l−FntFn+1l−Fn−1tFn+2t−FnkFn+1t−Fn−1ktt−kl−t.


### 5.3. On the Equationyn+1=yn−1−yn−1yn−3/yn−3+yn−5

In this section, we obtain the solution of the following special case of (1):(39)yn+1=yn−1−yn−1yn−3yn−3+yn−5,where the initial conditions y−5,y−4,y−3,y−2,y−1, and y0 are arbitrary positive real numbers.Theorem 6.
Letxnn=−5∞ be the solution of equation (39) satisfying y−5=k, y−4=h, y−3=t, y−2=s, y−1=l, and y0=r. Then, for n=0,1,2,…,(40)y4n−3=kltFn+1k+FntFn−1l+Fnt,y4n−2=hrsFn+1h+FnsFn−1r+Fns,y4n−1=kltFn+1k+FntFnl+Fn+1t,y4n=hrsFn+1h+FnsFnr+Fn+1s.Proof.
Forn=0, the result holds. Now, assume that n>0 and that our assumption holds for n−1, that is,(41)y4n−9=kltFn−1k+Fn−2tFn−2l+Fn−1ty4n−8=hrsFn−1h+Fn−2sFn−2r+Fn−1sy4n−7=kltFnk+Fn−1tFn−2l+Fn−1t,y4n−6=hrsFnh+Fn−1sFn−2r+Fn−1s,y4n−5=kltFnk+Fn−1tFn−1l+Fnt,y4n−4=hrsFnh+Fn−1sFn−1r+Fns.
It can be seen that(42)y4n−3=y4n−5−y4n−5y4n−7y4n−7+y4n−9=kltFnk+Fn−1tFn−1l+Fnt−klt/Fnk+Fn−1tFn−1l+Fntklt/Fnk+Fn−1tFn−2l+Fn−1tklt/Fnk+Fn−1tFn−2l+Fn−1t+klt/Fn−1k+Fn−2tFn−2l+Fn−1t=kltFnk+Fn−1tFn−1l+Fnt−klt/Fnk+Fn−1tFn−1l+Fntklt/Fnk+Fn−1tFn−2l+Fn−1tkltFn−1k+Fn−2t+Fnk+Fn−1t/FnK+Fn−1tFn−1K+Fn−2tFn−2l+Fn1−t=kltFnk+Fn−1tFn−1l+Fnt−kltFnk+Fn−1tFn−1l+FntFn−1k+Fn−2tFn−1k+Fn−2t+Fnk+Fn−1t=kltFn−1k+Fn−2t+Fnk+Fn−1t−kltFn−1k+Fn−2tFnk+Fn−1tFn−1l+FntFn−1k+Fn−2t+Fnk+Fn−1tFn−1l+FntFnk+Fn−1t=kltFnk+Fn−1tFnk+Fn−1tFn−1l+FntFn−1k+Fn−2t+Fnk+Fn−1tFn−1l+FntFnk+Fn−1t=kltFnk+Fn−1tFnk+Fn−1tFn−1l+FntFn−1k+Fn−2t+Fnk+Fn−1t=kltFn−1l+FntFn−1k+Fn−2t+Fnk+Fn−1t=kltFn+1k+FntFn−1l+Fnt.
Similarly,(43)y4n−1=y4n−3−y4n−3y4n−5y4n−5+y4n−7=kltFn+1k+FntFn−1l+Fnt−klt/Fn+1k+FntFn−1l+Fntklt/Fnk+Fn−1tFn−1l+Fntklt/Fnk+Fn−1tFn−1l+Fnt+klt/Fnk+Fn−1tFn−2l+Fn−1t=kltFn+1k+FntFn−1l+Fnt−klt/Fn+1k+FntFn−1l+Fntklt/Fnk+Fn−1tFn−1l+FntkltFn−2l+Fn−1t+Fn−1l+Fnt/Fnk+Fn−1tFn−1l+FntFn−2l+Fn−1t=kltFn+1k+FntFn−1l+Fnt−klt/Fn+1k+FntFn−1l+FntFn−2l+Fn−1t+Fn−1l+Fnt/Fn−2l+Fn−1t=kltFn+1k+FntFn−1l+Fnt−kltFn−2l+Fn−1tFn+1k+FntFn−1l+FntFn−2l+Fn−1t+Fn−1l+Fnt=kltFn−2l+Fn−1t+kltFn−1l+Fnt−kltFn−2l+Fn−1tFn+1k+FntFn−1l+FntFn−2l+Fn−1t+Fn+1k+FntFn−1l+FntFn−1l+Fnt=kltFn−1l+FntFn+1k+FntFn−1l+FntFn−2l+Fn−1t+Fn−1l+Fnt=kltFn+1k+FntFn−2l+Fn−1t+Fn−1l+Fnt=kltFn+1k+FntFnl+Fn+1t.


### 5.4. On the Equationyn+1=yn−1−yn−1yn−3/yn−3−yn−5

In this section, we present the periodicity of the solutions of the following equation:(44)yn+1=yn−1−yn−1yn−3yn−3−yn−5,where the initial conditions y−5,y−4,y−3,y−2,y−1, and y0 are arbitrary positive real numbers.Theorem 7.
Letxnn=−5∞ be a solution to equation (44). Then, every solution of equation (44) is periodic with period 12. Moreover, xnn=−5∞ takes the form(45)k,h,t,s,l,r,−lkt−k,−rhs−h,lktt−kl−t,rhss−hr−s,lkl−t,rhr−s,k,h,t,…,or(46)y12n−5=k,y12n−4=h,y12n−3=t,y12n−2=s,y12n−1=l,y12n=r,y12n+1=−lkt−k,y12n+2=−rhs−h,y12n+3=lktt−kl−t,y12n+4=rhss−hr−s,y12n+5=lkl−t,y12n+6=rhr−s,where y−5=k, y−4=h, y−3=t, y−2=s, y−1=l, and y0=r.Proof.
The proof is similar to the proof of the previous theorem. Hence, it will be omitted.


## 5.1. On the Equationyn+1=yn−1+yn−1yn−3/yn−3+yn−5

In this section, we present the exact solutions of equation (9):(28)yn+1=yn−1+yn−1yn−3yn−3+yn−5,where the initial conditions y−5,y−4,y−3,y−2,y−1, and y0 are arbitrary real numbers.Theorem 4.
Letxnn=−5∞ be a solution to equation (28) satisfying y−5=k, y−4=h, y−3=t, y−2=s, y−1=l, and y0=r. Then, for n=0,1,2,…,(29)y4n−3=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t,y4n−2=s∏i=0n−1F2i+2h+F2i+3sF2i+1r+F2isF2i+1h+F2i+2sF2ir+F2i−1s,y4n−1=l∏i=0n−1F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t,y4n=r∏i=0n−1F2i+2h+F2i+3sF2i+3r+F2i+2sF2i+1h+F2i+2sF2i+2r+F2i+1s,where Fmm=0∞=0,1,1,2.3,5,8,13,…Proof.
Forn=0, the result holds. Now, suppose that n>0 and that our assumption holds for n−1, that is,(30)y4n−9=l∏i=0n−3F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t,y4n−8=r∏i=0n−3F2i+2h+F2i+3sF2i+3r+F2i+2sF2i+1h+F2i+2sF2i+2r+F2i+1s,y4n−7=t∏i=0n−2F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t,y4n−6=s∏i=0n−2F2i+2h+F2i+3sF2i+1r+F2isF2i+1h+F2i+2sF2ir+F2i−1s,y4n−5=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t,y4n−4=r∏i=0n−2F2i+2h+F2i+3sF2i+3r+F2i+2sF2i+1h+F2i+2sF2i+2r+F2i+1s.
Now, it follows from equation (28) that(31)y4n−3=y4n−5+y4n−5y4n−7y4n−7+y4n−9=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t+l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2t/F2i+1k+F2i+2tF2i+2l+F2i+1tt∏i=0n−2F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1tt∏i=0n−2F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1t+l∏i=0n−3F2i+2k+F2i+3tF2i+3l+F2i+2t/F2i+1k+F2i+2tF2i+2l+F2i+1t=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t+l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2t/F2i+1k+F2i+2tF2i+2l+F2i+1tl∏i=0n−2F2i+2k+F2i+3t/F2i+1k+F2i+2tl∏i=0n−2F2i+2k+F2i+3t/F2i+1k+F2i+2t+l∏i=0n−3F2i+2k+F2i+3t/F2i+1k+F2i+2t=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t+l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2n−2k+F2n−1t/F2i+1k+F2i+2tF2i+2l+F2i+1tF2n−3k+F2n−2tF2n−2k+F2n−1t/F2n−3k+F2n−2t+1=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t1+F2n−2k+F2n−1t/F2n−3k+F2n−2tF2n−2k+F2n−1t/F2n−3k+F2n−2t+1=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t1+F2n−2k+F2n−1tF2n−2k+F2n−1t+F2n−3k+F2n−2t=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t1+F2n−2k+F2n−1tF2n−1k+F2nt=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1tF2n−2k+F2n−1t+F2n−1k+F2ntF2n−1k+F2nt=l∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1tF2nk+F2n+1tF2n−1k+F2nt.
Then,(32)y4n−3=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t.
Also, we get from equation (28) that(33)y4n−1=y4n−3+y4n−3y4n−5y4n−5+y4n−7=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1tl∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2t/F2i+1k+F2i+2tF2i+2l+F2i+1tl∏i=0n−2F2i+2k+F2i+3tF2i+3l+F2i+2t/F2i+1k+F2i+2tF2i+2l+F2i+1t+t∏i=0n−2F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1tl∏i=0n−2F2i+3l+F2i+2t/F2i+2l+F2i+1tl∏i=0n−2F2i+3l+F2i+2t/F2i+2l+F2i+1t+t∏i=0n−2F2i+1l+F2it/F2il+F2i−1t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+lt∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1t∏i=0n−2F2i+2l+F2i+1t/F2i+3l+F2i+2tl∏i=0n−2F2i+3l+F2i+2t/F2i+2l+F2i+1t+t∏i=0n−2F2i+1l+F2it/F2il+F2i−1t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+lt∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1tl+lF2n−2l+F2n−3t/F2n−1l+F2n−2t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1t1+F2n−2l+F2n−3t/F2n−1l+F2n−2t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1t1+F2n−2l+F2n−3t/F2n−1l+F2n−2tF2n−1l+F2n−2tF2n−1l+F2n−2t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1tF2n−1l+F2n−2tF2n−1l+F2n−2t+F2n−2l+F2n−3t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t+t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2it/F2i+1k+F2i+2tF2il+F2i−1tF2n−1l+F2n−2tF2nl+F2n−1t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1t1+F2n−1l+F2n−2tF2nl+F2n−1t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1tF2nl+F2n−1t+F2n−1l+F2n−2tF2nl+F2n−1t=t∏i=0n−1F2i+2k+F2i+3tF2i+1l+F2itF2i+1k+F2i+2tF2il+F2i−1tF2n+1l+F2ntF2nl+F2n−1t=l∏i=0n−1F2i+2k+F2i+3tF2i+3l+F2i+2tF2i+1k+F2i+2tF2i+2l+F2i+1t.
Thus, other relations can be similarly proved.


## 5.2. On the Equationyn+1=yn−1+yn−1yn−3/yn−3−yn−5

This section deals with the solutions of the following equation:(34)yn+1=yn−1+yn−1yn−3yn−3−yn−5,where the initial conditions y−5,y−4,y−3,y−2,y−1, and y0 are nonzero arbitrary real numbers.Theorem 5.
Letxnn=−5∞ be a solution to equation (34) satisfying y−5=k, y−4=h, y−3=t, y−2=s, y−1=l, and y0=r. Then, for n=0,1,2,…,(35)y4n−3=Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−Fnktt−kl−t,y4n−2=Fnr−Fn−2sFn+1r−Fn−1sFn+1s−Fn−1hFn+2s−Fnhsr−ss−h,y4n−1=Fn+2l−FntFn+1l−Fn−1tFn+2t−FnkFn+1t−Fn−1ktt−kl−t,y4n=Fn+2s−FnhFn+1s−Fn−1hFn+2r−FnsFn+1r−Fn−1ssr−ss−h.Proof.
Forn=0, the result holds. Now, suppose that n>0 and that our assumption holds for n−1, that is,(36)y4n−9=Fnl−Fn−2tFn−1l−Fn−3tFnt−Fn−2kFn−1t−Fn−3ktt−kl−t,y4n−8=Fns−Fn−2hFn−1s−Fn−3hFnr−Fn−2sFn−1r−Fn−3ssr−ss−h,y4n−7=Fn−1l−Fn−3tFnl−Fn−2tFnt−Fn−2kFn+1t−Fn−1ktt−kl−t,y4n−6=Fn−1r−Fn−3sFnr−Fn−2sFns−Fn−2hFn+1s−Fn−1hsr−ss−h,y4n−5=Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2ktt−kl−t,y4n−4=Fn+1s−Fn−1hFns−Fn−2hFn+1r−Fn−1sFnr−Fn−2ssr−ss−h.
Now, it can be seen from equation (34) that(37)y4n−3=y4n−5+y4n−5y4n−7y4n−7−y4n−9=Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2ktt−kl−t+Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2k/tt−kl−tFn−1l−Fn−3tFnl−Fn−2tFnt−Fn−2kFn+1t−Fn−1k/tt−kl−tFn−1l−Fn−3tFnl−Fn−2tFnt−Fn−2kFn+1t−Fn−1k/tt−kl−t−Fnl−Fn−2tFn−1l−Fn−3tFnt−Fn−2kFn−1t−Fn−3k/tt−kl−t=Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2ktt−kl−t+Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2k/tt−kl−tFn−1l−Fn−3tFnl−Fn−2tFnt−Fn−2kFn+1t−Fn−1k/tt−kl−tFn−1l−Fn−3tFnl−Fn−2tFnt−Fn−2k/tt−kl−tFn+1t−Fn−1k−Fn−1t−Fn−3k=Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2ktt−kl−t+Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2kFn+1t−Fn−1ktt−kl−tFn+1t−Fn−1k−Fn−1t−Fn−3k=Fn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2kFn+1t−Fn−1k2−Fn−1t−Fn−3ktt−kl−tFn+1t−Fn−1k−Fn−1t−Fn−3k=Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−Fnktt−kl−t.
It can also be observed that(38)y4n−1=y4n−3+y4n−3y4n−5y4n−5−y4n−7=Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−Fnktt−kl−t+Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−Fnk/tt−kl−tFn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2k/tt−kl−tFn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2k/tt−kl−tFn−1l−Fn−3tFnl−Fn−2tFnt−Fn−2kFn+1t−Fn−1k/tt−kl−t=Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−Fnktt−kl−t+Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−Fnk/tt−kl−tFn+1l−Fn−1tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2k/tt−kl−tFnl−Fn−2tFn+1t−Fn−1kFnt−Fn−2k/tt−kl−tFn+1l−Fn−1t−Fn−1l−Fn−3t=Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−Fnktt−kl−t+Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−FnkFn+1l−Fn−1ttt−kl−tFn+1l−Fn−1t−Fn−1l−Fn−3t=Fnl−Fn−2tFn+1l−Fn−1tFn+1t−Fn−1kFn+2t−FnkFn+1l−Fn−1t2−Fn−1l−Fn−3ttt−kl−tFn+1l−Fn−1t−Fn−1l−Fn−3t=Fn+2l−FntFn+1l−Fn−1tFn+2t−FnkFn+1t−Fn−1ktt−kl−t.


## 5.3. On the Equationyn+1=yn−1−yn−1yn−3/yn−3+yn−5

In this section, we obtain the solution of the following special case of (1):(39)yn+1=yn−1−yn−1yn−3yn−3+yn−5,where the initial conditions y−5,y−4,y−3,y−2,y−1, and y0 are arbitrary positive real numbers.Theorem 6.
Letxnn=−5∞ be the solution of equation (39) satisfying y−5=k, y−4=h, y−3=t, y−2=s, y−1=l, and y0=r. Then, for n=0,1,2,…,(40)y4n−3=kltFn+1k+FntFn−1l+Fnt,y4n−2=hrsFn+1h+FnsFn−1r+Fns,y4n−1=kltFn+1k+FntFnl+Fn+1t,y4n=hrsFn+1h+FnsFnr+Fn+1s.Proof.
Forn=0, the result holds. Now, assume that n>0 and that our assumption holds for n−1, that is,(41)y4n−9=kltFn−1k+Fn−2tFn−2l+Fn−1ty4n−8=hrsFn−1h+Fn−2sFn−2r+Fn−1sy4n−7=kltFnk+Fn−1tFn−2l+Fn−1t,y4n−6=hrsFnh+Fn−1sFn−2r+Fn−1s,y4n−5=kltFnk+Fn−1tFn−1l+Fnt,y4n−4=hrsFnh+Fn−1sFn−1r+Fns.
It can be seen that(42)y4n−3=y4n−5−y4n−5y4n−7y4n−7+y4n−9=kltFnk+Fn−1tFn−1l+Fnt−klt/Fnk+Fn−1tFn−1l+Fntklt/Fnk+Fn−1tFn−2l+Fn−1tklt/Fnk+Fn−1tFn−2l+Fn−1t+klt/Fn−1k+Fn−2tFn−2l+Fn−1t=kltFnk+Fn−1tFn−1l+Fnt−klt/Fnk+Fn−1tFn−1l+Fntklt/Fnk+Fn−1tFn−2l+Fn−1tkltFn−1k+Fn−2t+Fnk+Fn−1t/FnK+Fn−1tFn−1K+Fn−2tFn−2l+Fn1−t=kltFnk+Fn−1tFn−1l+Fnt−kltFnk+Fn−1tFn−1l+FntFn−1k+Fn−2tFn−1k+Fn−2t+Fnk+Fn−1t=kltFn−1k+Fn−2t+Fnk+Fn−1t−kltFn−1k+Fn−2tFnk+Fn−1tFn−1l+FntFn−1k+Fn−2t+Fnk+Fn−1tFn−1l+FntFnk+Fn−1t=kltFnk+Fn−1tFnk+Fn−1tFn−1l+FntFn−1k+Fn−2t+Fnk+Fn−1tFn−1l+FntFnk+Fn−1t=kltFnk+Fn−1tFnk+Fn−1tFn−1l+FntFn−1k+Fn−2t+Fnk+Fn−1t=kltFn−1l+FntFn−1k+Fn−2t+Fnk+Fn−1t=kltFn+1k+FntFn−1l+Fnt.
Similarly,(43)y4n−1=y4n−3−y4n−3y4n−5y4n−5+y4n−7=kltFn+1k+FntFn−1l+Fnt−klt/Fn+1k+FntFn−1l+Fntklt/Fnk+Fn−1tFn−1l+Fntklt/Fnk+Fn−1tFn−1l+Fnt+klt/Fnk+Fn−1tFn−2l+Fn−1t=kltFn+1k+FntFn−1l+Fnt−klt/Fn+1k+FntFn−1l+Fntklt/Fnk+Fn−1tFn−1l+FntkltFn−2l+Fn−1t+Fn−1l+Fnt/Fnk+Fn−1tFn−1l+FntFn−2l+Fn−1t=kltFn+1k+FntFn−1l+Fnt−klt/Fn+1k+FntFn−1l+FntFn−2l+Fn−1t+Fn−1l+Fnt/Fn−2l+Fn−1t=kltFn+1k+FntFn−1l+Fnt−kltFn−2l+Fn−1tFn+1k+FntFn−1l+FntFn−2l+Fn−1t+Fn−1l+Fnt=kltFn−2l+Fn−1t+kltFn−1l+Fnt−kltFn−2l+Fn−1tFn+1k+FntFn−1l+FntFn−2l+Fn−1t+Fn+1k+FntFn−1l+FntFn−1l+Fnt=kltFn−1l+FntFn+1k+FntFn−1l+FntFn−2l+Fn−1t+Fn−1l+Fnt=kltFn+1k+FntFn−2l+Fn−1t+Fn−1l+Fnt=kltFn+1k+FntFnl+Fn+1t.


## 5.4. On the Equationyn+1=yn−1−yn−1yn−3/yn−3−yn−5

In this section, we present the periodicity of the solutions of the following equation:(44)yn+1=yn−1−yn−1yn−3yn−3−yn−5,where the initial conditions y−5,y−4,y−3,y−2,y−1, and y0 are arbitrary positive real numbers.Theorem 7.
Letxnn=−5∞ be a solution to equation (44). Then, every solution of equation (44) is periodic with period 12. Moreover, xnn=−5∞ takes the form(45)k,h,t,s,l,r,−lkt−k,−rhs−h,lktt−kl−t,rhss−hr−s,lkl−t,rhr−s,k,h,t,…,or(46)y12n−5=k,y12n−4=h,y12n−3=t,y12n−2=s,y12n−1=l,y12n=r,y12n+1=−lkt−k,y12n+2=−rhs−h,y12n+3=lktt−kl−t,y12n+4=rhss−hr−s,y12n+5=lkl−t,y12n+6=rhr−s,where y−5=k, y−4=h, y−3=t, y−2=s, y−1=l, and y0=r.Proof.
The proof is similar to the proof of the previous theorem. Hence, it will be omitted.


## 6. Numerical Examples

Here, we will present some numerical examples to confirm the previous theoretical work.Example 1.
We plot the local stability of equation (9) under the values a=0.5,b=0.5,c=10,d=1,x−5=0.3,x−4=−1,x−3=1,x−2=0.5,x−1=−2, and x0=2 (see Figure 1).Figure 1
Local stability of equation (9).Example 2.
This example shows the global stability of equation (9) under the values a=0.5,b=0.2,c=1,d=1,x−5=0.5,x−4=1,x−3=−3,x−2=4,x−1=−9, and x0=8 (see Figure 2).Figure 2
Global stability of equation (9).Example 3.
Figure3 shows the bounded solution of equation (9) when we assume that a=0.25,b=2,c=5,d=1,x−5=−0.5,x−4=0.8,x−3=5,x−2=−3,x−1=4, and x0=1.Figure 3
Bounded solution of equation (9).Example 4.
This example shows the solution of equation (28) under the values a=1,b=1,c=1,d=1,x−5=0.5,x−4=−0.8,x−3=−0.6,x−2=0.2,x−1=2, and x0=−1 (see Figure 4).Figure 4
Solution of equation (28).Example 5.
In this example, we plot the solution of equation (34) when we take a=1,b=1,c=1,d=1,x−5=−0.5,x−4=0.8,x−3=−3,x−2=−1,x−1=2, and x0=−3, as can be seen in Figure 5.Figure 5
Solution of equation (34).Example 6.
The solution of equation (39) is presented in Figure 6. Here, we consider our values as follows: a=1,b=1,c=1,d=1,x−5=−2.7,x−4=2.3,x−3=−3,x−2=3,x−1=−1, and x0=2.Figure 6
Solution of equation (39).Example 7.
This example shows the periodicity of equation (44) under the values a=1,b=1,c=1,d=1,x−5=−2.7,x−4=2.3,x−3=−3,x−2=3,x−1=−1, and x0=2 (see Figure 7).Figure 7
Periodicity of equation (44).



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*Source: 1230979-2020-09-28.xml*