Subst ), we nave > Zn-h 1 kn+1 h=1 = ) kin-1 Vn | 2 EWn-p 1 = Vn-1. kn p=1 Vn+1 Wn - H Hence, we see that 1 1 k1 Z-1 + 2-2 Wo - H 20 - € ko w-1+ w_2 ko Vo = ond ||
Subst ), we nave > Zn-h 1 kn+1 h=1 = ) kin-1 Vn | 2 EWn-p 1 = Vn-1. kn p=1 Vn+1 Wn - H Hence, we see that 1 1 k1 Z-1 + 2-2 Wo - H 20 - € ko w-1+ w_2 ko Vo = ond ||
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 the determine yellow and all information is here
![Then, we assume that
Wn+1 - l
Wn - l
kn+1
kn =
1
E Wn-p
Wn-p
p=0
p=1
(6)
Zn+1 - €
Zn - €
Vn+1 =
Vn =
2
2 zn-h
> Zn-h
h=0
h=1
Substituting (6) in (5), we have
2
E Zn-h
1
kn-1)
h=1
kn+1
Zn - €
Vn
(7)
> Wn-p
1
= Vn-1.
km
Vn+1 =
Wn - H
Hence, we see that
1
v1 =
ko
Wo – l
20 - €
ko =
ki =
Vo
Vo =
W-1+ w_2
Z–1 + z_2
and
at n = 1, k2 = ko and
V2 = Vo,
at n = 2, k3 = k1 and
V3 = Vị,
at n = 3,
k4 = ko and
V4 = Vo,
at n = 4, kz = k, and
V5 = V1,
(8)
at n = 2 n, k2n
= ko and
V2n = Vo,
at n = 2 n + 1, k2n+1 = k, and
V2n+1 = V .
Now, from the relations in (6) we get
Wn =
H + kn (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 + k1 (w2n + W2n-1),
(10)
Z2n = € + Vo (Z2n-1+ 2n-2),
22n+1 = € +V1 (22n + %2n–-1).
||
||
||](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F158aed3a-44ed-4147-bb6c-9729d6a7101b%2Fbbdefcd9-3a46-4dc1-9403-6a3b4497a758%2F8lcv06t_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Then, we assume that
Wn+1 - l
Wn - l
kn+1
kn =
1
E Wn-p
Wn-p
p=0
p=1
(6)
Zn+1 - €
Zn - €
Vn+1 =
Vn =
2
2 zn-h
> Zn-h
h=0
h=1
Substituting (6) in (5), we have
2
E Zn-h
1
kn-1)
h=1
kn+1
Zn - €
Vn
(7)
> Wn-p
1
= Vn-1.
km
Vn+1 =
Wn - H
Hence, we see that
1
v1 =
ko
Wo – l
20 - €
ko =
ki =
Vo
Vo =
W-1+ w_2
Z–1 + z_2
and
at n = 1, k2 = ko and
V2 = Vo,
at n = 2, k3 = k1 and
V3 = Vị,
at n = 3,
k4 = ko and
V4 = Vo,
at n = 4, kz = k, and
V5 = V1,
(8)
at n = 2 n, k2n
= ko and
V2n = Vo,
at n = 2 n + 1, k2n+1 = k, and
V2n+1 = V .
Now, from the relations in (6) we get
Wn =
H + kn (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 + k1 (w2n + W2n-1),
(10)
Z2n = € + Vo (Z2n-1+ 2n-2),
22n+1 = € +V1 (22n + %2n–-1).
||
||
||
![In this paper, we solve and study the properties of the following system
E Wn-p
> Zn-h
Zn-h
Wn-P
p=0
h=1
p=1
h=0
Wn+1
+µ and zn+1
+ €,
(4)
Zn - €
Wn - H
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, zn}-
be a solution of (4), then
n=-2
o+ V02 – 4 ko k1
$ - V62 – 4 ko k1
k1 + ko
w2n-2
+ B
+
2
2
ko – k1
n
$ + Vø2 – 4 ko k1
$+ Vø2 – 4 ko k1
0 - V02 – 4 ko k1
0 - Vo2 – 4 ko k1
1
A
ko
1
+ B
ko
w2n-1
- 1
- 1
2
2
+ (-) (는)-)
(1– ko + k1
1– k1 – ko
1
- 1
ko
+ V2 – 4 vo vi
1 – vị + vo
+
- V62
4 vo v1
22n-2
+ D
E,
2
2
- vo - v1
1
+ V2 – 4 vọ v1
b + V2 – 4 vo v1
e;2 – 4 vo v1
al2 - 4 vo vị
22n-1
C
+ D
- 1
2
2
2
+(G-) )-) .
1- vi + vo
1- vo - v1
E,
where A, B, C and D are constants defined as
wo - u
z-1 + z-2
ko
ki =
O = ko + k1 + ko k1,
w-1+ w-2
20 - E
20 - €
w-1 + w-2
v1 =
= vo + vi + vo v1,
2-1+ z-2
1 – Om
Vo2 – 4 ko k1
2 62 – 8 ko ki
1– k1 + ko
1 –
1 – kị + ko`
1 — ко — k1
A =
- 4 ko k1 – o) + 2 ( wo –
w-2 -
V62 – 4 ko ki
2 ф2 — 8 kо k1
1 – ko + k1
- ko – k1
(1 – kị + ko) µ)|,
1 – k1 – ko
B
u - w-2
62 – 4 ko k1 + ¢) +2 ( wo –
=
4 vo v1
2 2 – 8 vo v1
1- vi + vo
1- vị + vo
2-2 -
1- vo - vi
1- vo - vi
V2 – 4 vo vi
1 – vị + vo
1- v1 + vo
E - z-2
2 – 4 vo v1
+
+2
20 -
2 2 – 8 vo vi
- vO - v1
1- vo - v1
since 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 follow
form
Zn-h
Wn-p
Wn+1 - H
h=1
Zn+1 - €
and
(5)
1
1
Zn - €
Wn - u
E Wn-p
> Zn-h
p=0
h=0
4
||
||](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F158aed3a-44ed-4147-bb6c-9729d6a7101b%2Fbbdefcd9-3a46-4dc1-9403-6a3b4497a758%2F0vpcjv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:In this paper, we solve and study the properties of the following system
E Wn-p
> Zn-h
Zn-h
Wn-P
p=0
h=1
p=1
h=0
Wn+1
+µ and zn+1
+ €,
(4)
Zn - €
Wn - H
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, zn}-
be a solution of (4), then
n=-2
o+ V02 – 4 ko k1
$ - V62 – 4 ko k1
k1 + ko
w2n-2
+ B
+
2
2
ko – k1
n
$ + Vø2 – 4 ko k1
$+ Vø2 – 4 ko k1
0 - V02 – 4 ko k1
0 - Vo2 – 4 ko k1
1
A
ko
1
+ B
ko
w2n-1
- 1
- 1
2
2
+ (-) (는)-)
(1– ko + k1
1– k1 – ko
1
- 1
ko
+ V2 – 4 vo vi
1 – vị + vo
+
- V62
4 vo v1
22n-2
+ D
E,
2
2
- vo - v1
1
+ V2 – 4 vọ v1
b + V2 – 4 vo v1
e;2 – 4 vo v1
al2 - 4 vo vị
22n-1
C
+ D
- 1
2
2
2
+(G-) )-) .
1- vi + vo
1- vo - v1
E,
where A, B, C and D are constants defined as
wo - u
z-1 + z-2
ko
ki =
O = ko + k1 + ko k1,
w-1+ w-2
20 - E
20 - €
w-1 + w-2
v1 =
= vo + vi + vo v1,
2-1+ z-2
1 – Om
Vo2 – 4 ko k1
2 62 – 8 ko ki
1– k1 + ko
1 –
1 – kị + ko`
1 — ко — k1
A =
- 4 ko k1 – o) + 2 ( wo –
w-2 -
V62 – 4 ko ki
2 ф2 — 8 kо k1
1 – ko + k1
- ko – k1
(1 – kị + ko) µ)|,
1 – k1 – ko
B
u - w-2
62 – 4 ko k1 + ¢) +2 ( wo –
=
4 vo v1
2 2 – 8 vo v1
1- vi + vo
1- vị + vo
2-2 -
1- vo - vi
1- vo - vi
V2 – 4 vo vi
1 – vị + vo
1- v1 + vo
E - z-2
2 – 4 vo v1
+
+2
20 -
2 2 – 8 vo vi
- vO - v1
1- vo - v1
since 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 follow
form
Zn-h
Wn-p
Wn+1 - H
h=1
Zn+1 - €
and
(5)
1
1
Zn - €
Wn - u
E Wn-p
> Zn-h
p=0
h=0
4
||
||
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