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C- -p - Vp² + 8pt2 212 Now, we calculate 72 and write in A1. Then we have p+ /p? – 4p (1 + 2p+ /4p + 1) 11 =

C- -p - Vp² + 8pt2 212 Now, we calculate 72 and write in A1. Then we have p+ /p? – 4p (1 + 2p+ /4p + 1) 11 =

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Show me the steps of deremine red and inf is here i need evey I need all the details step by step and inf is here

-p - Vp² + 8 pt2 272 Now, we calculate i2 and write in 21. Then we have р2 — 4p (1 + 2р + /4p +1) 1+2p + /4p + 1 p+v-7p2 – 4p – 4p/I+4p 1+2p + /4p + p + λ p+ /7p? + 4p +4p/T+4pi 1+2p+ /4p + 1 V Thus straightforward calculations show that 2/2p 1+ I+4p Additionally, we obtain similarly calculations that 2./2p 1+ VI+4p %3D On the other hand, we consider 23 as follows: -p+ v9p? +4p+4p/4p+1 13 1+2p+ /4p +1 2 -p+V (3p+ T+ 4p) 1+2p + /4p +1 1– 2p/4p+1 2. -p+V (3p+ /I+4p)* 1+ 2p + /4p +1 2p+ I+4p < 1. 1+2p+ /4p+1 Moreover we clearly see that 23 > 0. So0 < 13 < 1 for all p > 0. Similar calculations we have that -1 < ^4 < O for all p > 0.
Transcribed Image Text:-p - Vp² + 8 pt2 272 Now, we calculate i2 and write in 21. Then we have р2 — 4p (1 + 2р + /4p +1) 1+2p + /4p + 1 p+v-7p2 – 4p – 4p/I+4p 1+2p + /4p + p + λ p+ /7p? + 4p +4p/T+4pi 1+2p+ /4p + 1 V Thus straightforward calculations show that 2/2p 1+ I+4p Additionally, we obtain similarly calculations that 2./2p 1+ VI+4p %3D On the other hand, we consider 23 as follows: -p+ v9p? +4p+4p/4p+1 13 1+2p+ /4p +1 2 -p+V (3p+ T+ 4p) 1+2p + /4p +1 1– 2p/4p+1 2. -p+V (3p+ /I+4p)* 1+ 2p + /4p +1 2p+ I+4p < 1. 1+2p+ /4p+1 Moreover we clearly see that 23 > 0. So0 < 13 < 1 for all p > 0. Similar calculations we have that -1 < ^4 < O for all p > 0.
Motivated by difference equations and their systems, we consider the following system of difference equations Yn , Yn+1 = A + B- Yn-1 Xn+1 = A + B- 2 (1) where A and B are positive numbers and the initial values are positive numbers. In First of all, we consider the change of the variables for system (1) as follows: Xn , Zn = A Уп tn = A From this, system (1) transform into following system: tn Zn , Zn+1 = 1+P2 Zn-1 tn+1 = 1+P2 (2) %3| 'n-1 where p = > 0. From now on, we study the system (2). A2 Lemma 1 Let p > 0. Unique positive equilibrium point of system (2) is (1+ I+4p 1+ VI+4p (F, 2) = ( - 2 2 Now, 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-1 2n-1 matrix about equilibrium point (7, z): 1 B (,३) = 12 0 0 1 Thus, the linearized system of system (2) about the unique positive equilibrium point is given by XN+1 B (i, 2) XN, where In tn-1 XN = Zn Zn-1 1 B (ī, 2): 0 0 1 Hence, the characteristic equation of B (i, z) about the unique positive equilibrium point (7, z) is 4p2 + 4p2 = 0. 14 Due to t = 7, we can rearrange the characteristic equation such that ²+ 4p², 4p2 74 14 = 0. Therefore, we obtain the four roots of characteristic equation as follows: p+ Vp? – 8pi? 272 – Vp² – 8pi? 212 p- V 12 = -p+/p? + 8pi? 272 13 %3D Fo o o
Transcribed Image Text:Motivated by difference equations and their systems, we consider the following system of difference equations Yn , Yn+1 = A + B- Yn-1 Xn+1 = A + B- 2 (1) where A and B are positive numbers and the initial values are positive numbers. In First of all, we consider the change of the variables for system (1) as follows: Xn , Zn = A Уп tn = A From this, system (1) transform into following system: tn Zn , Zn+1 = 1+P2 Zn-1 tn+1 = 1+P2 (2) %3| 'n-1 where p = > 0. From now on, we study the system (2). A2 Lemma 1 Let p > 0. Unique positive equilibrium point of system (2) is (1+ I+4p 1+ VI+4p (F, 2) = ( - 2 2 Now, 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-1 2n-1 matrix about equilibrium point (7, z): 1 B (,३) = 12 0 0 1 Thus, the linearized system of system (2) about the unique positive equilibrium point is given by XN+1 B (i, 2) XN, where In tn-1 XN = Zn Zn-1 1 B (ī, 2): 0 0 1 Hence, the characteristic equation of B (i, z) about the unique positive equilibrium point (7, z) is 4p2 + 4p2 = 0. 14 Due to t = 7, we can rearrange the characteristic equation such that ²+ 4p², 4p2 74 14 = 0. Therefore, we obtain the four roots of characteristic equation as follows: p+ Vp? – 8pi? 272 – Vp² – 8pi? 212 p- V 12 = -p+/p? + 8pi? 272 13 %3D Fo o o
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