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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
show me steps of the determine blue and all information is here explain to me step by step i need every details

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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F158aed3a-44ed-4147-bb6c-9729d6a7101b%2F84ecfada-6c30-4ad6-8817-52ab0a9650e5%2Fziwqrhu_processed.png&w=3840&q=75)
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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