=0 Πιο Ω_q+1)t-q + 1 ΠΩ(q+1)t-q Πιο Ω_q+1)t-q + 1 ΠoΩ(q+1)t-q/Ω(64+5) thus Ω(7q+6) > Ω—(6q+5)• Th h=0 k=1 1 Πιο Ω(q+1)k-(q+1)t-q + 1 o Th+1 1 -ΣΠ Πιο Ω(q+1)k-(q+1)t-q +1 h=0 k=1
=0 Πιο Ω_q+1)t-q + 1 ΠΩ(q+1)t-q Πιο Ω_q+1)t-q + 1 ΠoΩ(q+1)t-q/Ω(64+5) thus Ω(7q+6) > Ω—(6q+5)• Th h=0 k=1 1 Πιο Ω(q+1)k-(q+1)t-q + 1 o Th+1 1 -ΣΠ Πιο Ω(q+1)k-(q+1)t-q +1 h=0 k=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 the steps of determine blue and all information is here
Transcribed Image Text:From the equations (3) and (4),
C.
7h
II-o 2-(q+1)t-q+1
TI-, 2-(q+1)t-q
ΣΠ
1
t30
2IIT g+1)k-(q+1)t-q † 1
>
t=0
h=0 k=1
0o 7h+1
II-0 2-(q+1)t-q
II-o 2-(4+1)t-q/S2-(6q+5)
+ 1
1
ΣΠ
II-o P(a+1)k-(q+1)t-q +1
t=0
h=0 k=1
=0
thus 2-(79+6) > S2-(6q+5).
Transcribed Image Text:In this work, we deal with the following nonlinear difference equation
Im-(7g+6)
1+II-o Pm-(a+1)t-g
2-1, lo E (0, 0) is investigated.
Sm+1
m = 0, 1, ..,
(1)
t%3D0
where 2-(79+6), S2-(79+5),
(d) We can generate the following formulas:
L(79+7)n+r(q+1)+s+1 = (r-7)(4+1)+s+1 (1–
(IIm=1 2-(mg+m-1)+s)/l(r-7)(g+1)+s+1
1+ (IIm=1
-D1
-(mg+m-1)+s)
7h+r
1
ΣΠ
Ilm-1 P(a+1)t-(mg+m-1)+s + 1
h=0 k=1
=D1
(e) If S2(7g+7)n+(t-1)q+t → a(t-1)q+t 70 then 279+7)n+6q+7 → 0 as n → 0,
If 2(79+7)n+tq+t → atg+t +0 then N(7g+7)n+79+7 + 0 as n → 0. t = 1,6.
(e) Suppose that a1 = ag+2 = a2q+3 = a3q+4
= a4g+5
a5q+6
A6q+7
0. By (d), we
produce the following formulas below:
lim N(79+7)n+1
lim 2-(79+6)
1
II-o 2-(q+1)t-9
-(q+1)t-q
+1
It%3D0
n
7h
ΣΠ
1
LII TE 2(a+1)k-(q+1)t-q †
h=0 k=1
Lt30
T, N-(9+1)t-q5?-114921I ,(a+1)k-(4+1)t-9
7h
1
=N-(79+6)
1
II-o 2-(9+1)t-g+1
+1
h=0 k=1
It=D0
7h
II-o 2-(9+1)t-q+1
II-0 2-(a+1)t-9
1
EIIE g+1)k-(q+1)t-q ™ +
aj = 0 >
(3)
=
h=0 k=1
t%3D0
II-o 2-(4+1)t-q/2-(6q+5).
1
lim N(79+7)n+q+2= lim 2-(6q+5)
II-o 2-(9+1)t-q+1
n00
3D0
7h+1
1
ΣΠ
II-, 2-(9+1)k-(q+1)t-q
+
h=0 k=1
It=0
II-, 2-(9+1)t-q/2-(6q+5)
II-0 2-(a+1)t-q +1
aq+2
N-(64+5)
1
0o 7h+1
1
ΣΠ
-(a+1)k-(q+1)t-q
+1
h=0 k=1
00 7h+1
II-, 2-(4+1)t-g
II-,2-(4+1)t-g/2-(6q+5)
+1
1
-ΣΠ
t=0
(4)
II-o Pa+1)k-(9+1)t-g +1
ag+2 = 0 =
=0
h=0 k=1
%3D0
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