Show me the determine red and all information is here Let w2n-1 = Un, W2n-2 = Vn, 22n–1 = Xn and z2n-2 = Yn, implies thatVn+1 – ko Un – kio Vn = µ,(11)Un+1 – kį Vn+1 – kị Un = µ,andYn+1 – Vo Xn – v Yn = €,(12)Xn+1 – Vị Yn+1 – V1 Xn = €.It is observe that, the equations (11) and (12) are system of linear non-homogeneous differenceequations with constants coefficients. First, we solve the system in (11). Thus, consider (11) is(E – ko) Vn – ko Un(13)(E – ki) Un – ki E V,(14)where E = A + 1, is the shift operator. Multiplying (13) and (14) by (E – k1) and ko,respectively,(E – k1) (E – ko) Vn – (E – k1) ko Un(E – ki) µ,(15)ko (E – k1) Un – ko ki E V,ko fH.(16)Thus, the combination of (15) and (16) is(E²(ko + ki + ko ki) E + ko k1) Vn(1 – k1 + ko) µ,(17)(E? – o E + ko ki) Vn = (1 – k1 + ko) µ,where ø = ko + k1 + ko k1. Clearly, (17) is linear difference equations from order two and it iseasy to obtain the solution,nø + Vo? – 4 ko kı6² – 4 ko k'1(1– kị + ko1 – ko – k1.Ф —-Vn= A+ B(18)2since A and B are constants. By Substituting (18) in (13), we give1Un = Ako- 4 ko k1ф +14 ko ki226² – 4 ko k10² – 4 ko kiФ —+ B1(19)2(G-)(1– k1 + ko+1 — kо — ki,H.ko6. In this paper, we solve and study the properties of the following systemE Wn-p> Zn-hZn-hWn-Pp=0h=1p=1h=0Wn+1+µ and zn+1+ €,(4)Zn - €Wn - Hwhere u and e are arbitrary positive real numbers with initial conditions w; and z; for i =-2, –1,0.Theorem 2.1. Let {wn, zn}-be a solution of (4), thenn=-2o+ V02 – 4 ko k1$ - V62 – 4 ko k1k1 + kow2n-2+ B+22ko – k1n$ + Vø2 – 4 ko k1$+ Vø2 – 4 ko k10 - V02 – 4 ko k10 - Vo2 – 4 ko k11Ako1+ Bkow2n-1- 1- 122+ (-) (는)-)(1– ko + k11– k1 – ko1- 1ko+ V2 – 4 vo vi1 – vị + vo+- V624 vo v122n-2+ DE,22- vo - v11+ V2 – 4 vọ v1b + V2 – 4 vo v1e;2 – 4 vo v1al2 - 4 vo vị22n-1C+ D- 1222+(G-) )-) .1- vi + vo1- vo - v1E,where A, B, C and D are constants defined aswo - uz-1 + z-2koki =O = ko + k1 + ko k1,w-1+ w-220 - E20 - €w-1 + w-2v1 == vo + vi + vo v1,2-1+ z-21 – OmVo2 – 4 ko k12 62 – 8 ko ki1– k1 + ko1 –1 – kị + ko`1 — ко — k1A =- 4 ko k1 – o) + 2 ( wo –w-2 -V62 – 4 ko ki2 ф2 — 8 kо k11 – ko + k1- ko – k1(1 – kị + ko) µ)|,1 – k1 – koBu - w-262 – 4 ko k1 + ¢) +2 ( wo –=4 vo v12 2 – 8 vo v11- vi + vo1- vị + vo2-2 -1- vo - vi1- vo - viV2 – 4 vo vi1 – vị + vo1- v1 + voE - z-22 – 4 vo v1++220 -2 2 – 8 vo vi- vO - v11- vo - v1since ko + ki 1 and vo +v1 71 for n € N.Proof. To obtain the expressions of the general solutions for (4), we rewrite it in the followformZn-hWn-pWn+1 - Hh=1Zn+1 - €and(5)11Zn - €Wn - uE Wn-p> Zn-hp=0h=04||||

Answered: Image /qna-images/answer/811504de-04e5-4ae6-b6d9-567a09bda131.jpg

(E – ko) Vn – ko Un (13) (E – ki) U, – kị E V, (14) where E = A + 1, is the shift operator. Multiplying (13) and (14) by (E – k1) and ko, respectively, (E – k)) (E – ko) Vn – (E – k1) ko Un (E – k) µ, (15) ko (E – k1) Un – ko ki E V, ko H. (16) Thus, the combination of (15) and (16) is (E – (ko + ki + ko ki) E + ko k1) Vn = (1 – kı + ko) H, (E² – 6 E + ko ki) Vn (17) (1 – k1 + ko) H, where o = ko + ki + ko k1. Clearly, (17) is linear difference equations from order two and it is easy to obtain the solution, 4² – 4 ko k1 $- V6? – 4 ko ki 1- k1 + ko + 1- ko – k1 Vn = A + B (18) 2 since A and B are constants. By Substituting (18) in (13), we give n 0 + V02 – 4 ko ki ko $ + V02 - 4 ko ki Un = A 2 2 1 + B ko 62 – 4 ko k1 - 1 42 – 4 ko ki ф — (19) 2

(E – ko) Vn – ko Un (13) (E – ki) U, – kị E V, (14) where E = A + 1, is the shift operator. Multiplying (13) and (14) by (E – k1) and ko, respectively, (E – k)) (E – ko) Vn – (E – k1) ko Un (E – k) µ, (15) ko (E – k1) Un – ko ki E V, ko H. (16) Thus, the combination of (15) and (16) is (E – (ko + ki + ko ki) E + ko k1) Vn = (1 – kı + ko) H, (E² – 6 E + ko ki) Vn (17) (1 – k1 + ko) H, where o = ko + ki + ko k1. Clearly, (17) is linear difference equations from order two and it is easy to obtain the solution, 4² – 4 ko k1 $- V6? – 4 ko ki 1- k1 + ko + 1- ko – k1 Vn = A + B (18) 2 since A and B are constants. By Substituting (18) in (13), we give n 0 + V02 – 4 ko ki ko $ + V02 - 4 ko ki Un = A 2 2 1 + B ko 62 – 4 ko k1 - 1 42 – 4 ko ki ф — (19) 2

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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