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Now, from the relations in (6) we get Wn = µ+ kn (Wn-1 + Wn-2) and zn = € + Vn (žn-1 + Zn-2). (9) Using (8) in (9), we have W2n = H + ko (W2n-1+ W2n-2), W2n+1 = l+ k1 (w2n + w2n-1), (10) Z2n = € + vo (z2n-1+ 22n-2), Z2n+1 = €+ v1 (22n + 22n-1).

Now, from the relations in (6) we get Wn = µ+ kn (Wn-1 + Wn-2) and zn = € + Vn (žn-1 + Zn-2). (9) Using (8) in (9), we have W2n = H + ko (W2n-1+ W2n-2), W2n+1 = l+ k1 (w2n + w2n-1), (10) Z2n = € + vo (z2n-1+ 22n-2), Z2n+1 = €+ v1 (22n + 22n-1).

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 steps of the determine blue and all information is here explain to me step by step i need every details

Then, we assume that Wn+1 - l Wn - H kin+1 kn = 2 1 > Wn-p > Wn-p p=0 p=1 (6) Zn+1 - € Zn - € Vn+1 Vn = 1 E Zn-h h=0 E Zn-h h=1 Substituting (6) in (5), we have Zn-h 1 = kn-1) Vn h=1 kn+1 (7) 2 > Wn-p 1 = Vn-1· kn p=1 Vn+1 Wn - l Hence, we see that wo - H 20 - € 1 1 ko W_1+ w_2 Vo = Z–1 + z_2 Vo ko and at n = 1, k2 = ko and V2 = Vo, at n = 2, k3 = k1 and V3 = V1, at n = 3, k4 = ko and V4 = Vo, at n = 4, k5 = k1 and V5 = V1, (8) at n = 2 n, k2n = ko and V2n = Vo, at n = 2n + 1, k2n+1 = k1 and V2n+1 = V1. Now, from the relations in (6) we get Wn = u+ kin (Wn-1 + Wn-2) and zn = € + Vn (žn–1 + Zn-2). (9) Using (8) in (9), we have W2n = µ + ko (w2n–1 + W2n-2), W2n+1 = µ + kı (W2n + W2n-1), (10) 22n = € + vo (Z2n-1+ 2n-2), 22n+1 = € + V1 (22n + 2n–1). 5 .. || ||
Transcribed Image Text:Then, we assume that Wn+1 - l Wn - H kin+1 kn = 2 1 > Wn-p > Wn-p p=0 p=1 (6) Zn+1 - € Zn - € Vn+1 Vn = 1 E Zn-h h=0 E Zn-h h=1 Substituting (6) in (5), we have Zn-h 1 = kn-1) Vn h=1 kn+1 (7) 2 > Wn-p 1 = Vn-1· kn p=1 Vn+1 Wn - l Hence, we see that wo - H 20 - € 1 1 ko W_1+ w_2 Vo = Z–1 + z_2 Vo ko and at n = 1, k2 = ko and V2 = Vo, at n = 2, k3 = k1 and V3 = V1, at n = 3, k4 = ko and V4 = Vo, at n = 4, k5 = k1 and V5 = V1, (8) at n = 2 n, k2n = ko and V2n = Vo, at n = 2n + 1, k2n+1 = k1 and V2n+1 = V1. Now, from the relations in (6) we get Wn = u+ kin (Wn-1 + Wn-2) and zn = € + Vn (žn–1 + Zn-2). (9) Using (8) in (9), we have W2n = µ + ko (w2n–1 + W2n-2), W2n+1 = µ + kı (W2n + W2n-1), (10) 22n = € + vo (Z2n-1+ 2n-2), 22n+1 = € + V1 (22n + 2n–1). 5 .. || ||
In this paper, we solve and study the properties of the following system Wn-p 2 zn-h p=0 E wn-p zZn-h h=1 h=0 Wn+1 = p=1 + u and zn+1 = + €, (4) Zn - € Wn - u where u and e are arbitrary positive real numbers with initial conditions w; and z; for i = -2, –1,0. Theorem 2.1. Let {wn, z,} 2 be a solution of (4), then + V2 – 4 ko kı A - 4 ko ki - kị + ko w2n-2 +B ko ki + V – 4 ko kị $+ v2 – 4 ko ki - V – 4 ko kı 1 +B A ko w2n-1 1- ko + V2 – 4 vo VỊ V2 – 4 vo V - V + vo 22n-2 + D 0 - V1 1 (+ Ve2 – 4 v0 vi + V2 – 4 vo vi gb - V2 – 4 0 vỊ V2 – 4 vo v1 22n-1 +D 2 1- v + vo 1- vo - VI where A, B, C and D are constants defined as ko wo -4 ki = z-1 +z-2 $ = ko + ki + ko k1, w-1 +w-2 20 -e 20 -e w-1 + w-2 = vo+ v1 + o V, z-1 +2-2 wo -H V2 – 4 ko ki 262 – 8 ko ki 4) (V2 – 4 ko kı – 4) +2 A w-2- - 1- ko - ki V2 – 4 ko ki 2 62 - 8 ko ki - ko + k1 (1– ki + ko --) (v - sho ki +e) + 2 (wo - ( ) )]. B 4- w-2 - ko - ki 1- k - ko V2 – 4 vo vI 2 2 - 8 v0 v - vi + vo - vO - V , - v + o 1- vo - V +2 20 V2 – 4 v0 v1 1- v + vo 1-v + vo D E- 2-2 12 - 4 vO +2 2 2 since ko + ki #1 and vo + v 71 for n e N. Proof. To obtain the expressions of the general solutions for (4), we rewrite it in the follow form Zn-h Wn-p Wn+1 - 4 h=1 Zn+1 - € p=1 and (5) 2, - € Wn -H EWn-p 2 Zn-h h=0 p=0 Then, we assume that Wn+1 - u Wn - u kn+1 = 1 k, = Wn-p Wn-p p=0 p=1 (6) Zn+1 - € Zn - € Vn+1 = 1 Vn = 2 E zn-h E zn-h h=0 h=1 Substituting (6) in (5), we have 2 Zn-h h=1 kn+1 = 1 = -= kn-1, Vn (7) Wn-p p=1 1 = Vn-1. kn Vn+1 %3D Wn - u Hence, we see that 1 v1 = Vo wo - u 20 - € 1 ko = Vo = k = w-1 + w-2 Z-1 + 2-2 ko and at n = 1, k2 = ko and V2 = Vo, at n = 2, k3 = k1 and V3 = V1, at n = 3, k4 = ko and V4 = Vo, at n = 4, kg = k1 and V5 = V1, (8) at n = 2 n, k2n = ko and V2n = Vo, at n = 2n + 1, k2n+1 = k1 and V2n+1 = V1. Now, from the relations in (6) we get Wn = µ+ kn (wn-1 + Wn-2) and zn = €++ Vn (Zn-1 + zn-2). (9) Using (8) in (9), we have w2n = µ+ ko (w2n-1 + W2n-2), W2n+1 = H + k, (w2n + w2n-1), (10) 22n = € + vo (z2n-1+ 22n-2), Z2n+1 = €+ v1 (22n + z2n-1). 介
Transcribed Image Text:In this paper, we solve and study the properties of the following system Wn-p 2 zn-h p=0 E wn-p zZn-h h=1 h=0 Wn+1 = p=1 + u and zn+1 = + €, (4) Zn - € Wn - u where u and e are arbitrary positive real numbers with initial conditions w; and z; for i = -2, –1,0. Theorem 2.1. Let {wn, z,} 2 be a solution of (4), then + V2 – 4 ko kı A - 4 ko ki - kị + ko w2n-2 +B ko ki + V – 4 ko kị $+ v2 – 4 ko ki - V – 4 ko kı 1 +B A ko w2n-1 1- ko + V2 – 4 vo VỊ V2 – 4 vo V - V + vo 22n-2 + D 0 - V1 1 (+ Ve2 – 4 v0 vi + V2 – 4 vo vi gb - V2 – 4 0 vỊ V2 – 4 vo v1 22n-1 +D 2 1- v + vo 1- vo - VI where A, B, C and D are constants defined as ko wo -4 ki = z-1 +z-2 $ = ko + ki + ko k1, w-1 +w-2 20 -e 20 -e w-1 + w-2 = vo+ v1 + o V, z-1 +2-2 wo -H V2 – 4 ko ki 262 – 8 ko ki 4) (V2 – 4 ko kı – 4) +2 A w-2- - 1- ko - ki V2 – 4 ko ki 2 62 - 8 ko ki - ko + k1 (1– ki + ko --) (v - sho ki +e) + 2 (wo - ( ) )]. B 4- w-2 - ko - ki 1- k - ko V2 – 4 vo vI 2 2 - 8 v0 v - vi + vo - vO - V , - v + o 1- vo - V +2 20 V2 – 4 v0 v1 1- v + vo 1-v + vo D E- 2-2 12 - 4 vO +2 2 2 since ko + ki #1 and vo + v 71 for n e N. Proof. To obtain the expressions of the general solutions for (4), we rewrite it in the follow form Zn-h Wn-p Wn+1 - 4 h=1 Zn+1 - € p=1 and (5) 2, - € Wn -H EWn-p 2 Zn-h h=0 p=0 Then, we assume that Wn+1 - u Wn - u kn+1 = 1 k, = Wn-p Wn-p p=0 p=1 (6) Zn+1 - € Zn - € Vn+1 = 1 Vn = 2 E zn-h E zn-h h=0 h=1 Substituting (6) in (5), we have 2 Zn-h h=1 kn+1 = 1 = -= kn-1, Vn (7) Wn-p p=1 1 = Vn-1. kn Vn+1 %3D Wn - u Hence, we see that 1 v1 = Vo wo - u 20 - € 1 ko = Vo = k = w-1 + w-2 Z-1 + 2-2 ko and at n = 1, k2 = ko and V2 = Vo, at n = 2, k3 = k1 and V3 = V1, at n = 3, k4 = ko and V4 = Vo, at n = 4, kg = k1 and V5 = V1, (8) at n = 2 n, k2n = ko and V2n = Vo, at n = 2n + 1, k2n+1 = k1 and V2n+1 = V1. Now, from the relations in (6) we get Wn = µ+ kn (wn-1 + Wn-2) and zn = €++ Vn (Zn-1 + zn-2). (9) Using (8) in (9), we have w2n = µ+ ko (w2n-1 + W2n-2), W2n+1 = H + k, (w2n + w2n-1), (10) 22n = € + vo (z2n-1+ 22n-2), Z2n+1 = €+ v1 (22n + z2n-1). 介
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