Show me the steps of deremine yellow and inf is here i need evey I need all the details step by step and inf is here. It complete —р — у- Vp² + 8pr²14 =212Now, we calculate 72 and write in 11. Then we havep+ Vp? – 4p (1 + 2p+ /4p+ 1)1+2p + /4p +1p + v-7p² – 4p –- 4p/I+4p1+2p + /4p+ 1p+ /7p? + 4p + 4p/T+4pi1+2p+ /4p +1Thus straightforward calculations show that2/2p1+ /I+4pAdditionally, we obtain similarly calculations that2./2p|^2| =1+ /1+4pOn the other hand, we consider 13 as follows:-p+ v9p2 +4p+4p/4p+1131+2p+ V4p +1-p+V (3p + T+4p)ʻ – 1 – 2p 4p+I1+ 2p + 4p +12.-p+V(3p+ /T+4p)²1+ 2p + V4p +12р + +4p 0. So0 0. Unique positive equilibrium point of system (2) is(1+ I+4p 1+ VI+4p(F, 2) = ( -22Now, we consider a transformation as follows:(tn, tn-1, Zn, Zn-1) →(f, fi, 8, 81)where f = 1+ p, fi = tn, g = 1 + p, 81 = Zn. Thus we get the jacobian'n-12n-1matrix about equilibrium point (7, z):1B (,३) =120 01Thus, the linearized system of system (2) about the unique positive equilibrium pointis given by XN+1B (i, 2) XN, whereIntn-1XN =ZnZn-11B (ī, 2):0 01Hence, the characteristic equation of B (i, z) about the unique positive equilibriumpoint (7, z) is4p2+4p2= 0.14Due to t = 7, we can rearrange the characteristic equation such that²+ 4p², 4p27414= 0.Therefore, we obtain the four roots of characteristic equation as follows:p+ Vp? – 8pi?272– Vp² – 8pi?212p- V12 =-p+/p? + 8pi?27213%3DFo o o

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Show me the steps of deremine yellow and inf is here i need evey I need all the details step by step and inf is here. It complete —р — у- Vp² + 8pr²14 =212Now, we calculate 72 and write in 11. Then we havep+ Vp? – 4p (1 + 2p+ /4p+ 1)1+2p + /4p +1p + v-7p² – 4p –- 4p/I+4p1+2p + /4p+ 1p+ /7p? + 4p + 4p/T+4pi1+2p+ /4p +1Thus straightforward calculations show that2/2p1+ /I+4pAdditionally, we obtain similarly calculations that2./2p|^2| =1+ /1+4pOn the other hand, we consider 13 as follows:-p+ v9p2 +4p+4p/4p+1131+2p+ V4p +1-p+V (3p + T+4p)ʻ – 1 – 2p 4p+I1+ 2p + 4p +12.-p+V(3p+ /T+4p)²1+ 2p + V4p +12р + +4p< 1.1+2p + 4p+1Moreover we clearly see that 23 > 0. So0 < 13 < 1 for all p > 0. Similar calculationswe have that -1 < 4 < 0 for all p > 0. Motivated by difference equations and their systems, we consider the followingsystem of difference equationsYn, Yn+1 = A + B-Yn-1Xn+1 = A + B-2(1)where A and B are positive numbers and the initial values are positive numbers. InFirst of all, we consider the change of the variables for system (1) as follows:Xn, Zn =AУпtn =AFrom this, system (1) transform into following system:tnZn, Zn+1 = 1+P2Zn-1tn+1 = 1+P2(2)%3|'n-1where p = > 0. From now on, we study the system (2).A2Lemma 1 Let p > 0. Unique positive equilibrium point of system (2) is(1+ I+4p 1+ VI+4p(F, 2) = ( -22Now, we consider a transformation as follows:(tn, tn-1, Zn, Zn-1) →(f, fi, 8, 81)where f = 1+ p, fi = tn, g = 1 + p, 81 = Zn. Thus we get the jacobian'n-12n-1matrix about equilibrium point (7, z):1B (,३) =120 01Thus, the linearized system of system (2) about the unique positive equilibrium pointis given by XN+1B (i, 2) XN, whereIntn-1XN =ZnZn-11B (ī, 2):0 01Hence, the characteristic equation of B (i, z) about the unique positive equilibriumpoint (7, z) is4p2+4p2= 0.14Due to t = 7, we can rearrange the characteristic equation such that²+ 4p², 4p27414= 0.Therefore, we obtain the four roots of characteristic equation as follows:p+ Vp? – 8pi?272– Vp² – 8pi?212p- V12 =-p+/p? + 8pi?27213%3DFo o o

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On the other hand, we consider 23 as follows: -p+v9p? +4p+4p/4p+1 1+2p+ /4p+I 13 -p+V(3p + /T+4p)ʻ – 1 – 2p/4p+1 1+2p + /4p +1 -p+V(3p+ +4p)* 1+2p+ /4p+1 2p+ I+4p < 1. 1+2p + /4p+1