Πο Ω(q+1)t-q + 1 ΠoΩ_q+1)t-q/Ω(6q+5) ΠoΩ (q+1)t-q + 1 ΠoΩ(q+1)t-q/Ω(5q+4) o Th+1 1 = -ΣΠ h=o k=1 Πο Ω(q+1)k-(q+1)t-q| +1 thus Ω(6q+5) > Ω-(5q+4). = o Th+2 ΣΠ h=0 k=1 1 Πιο Ω(q+1)k-(a+1)t-q +1
Πο Ω(q+1)t-q + 1 ΠoΩ_q+1)t-q/Ω(6q+5) ΠoΩ (q+1)t-q + 1 ΠoΩ(q+1)t-q/Ω(5q+4) o Th+1 1 = -ΣΠ h=o k=1 Πο Ω(q+1)k-(q+1)t-q| +1 thus Ω(6q+5) > Ω-(5q+4). = o Th+2 ΣΠ h=0 k=1 1 Πιο Ω(q+1)k-(a+1)t-q +1
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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Question
show me the steps of determine blue and information is here
Transcribed Image Text:00 7h+5
II, 2-(a+1)t–q + 1
II-, 2-(4+1)t-q/2-(2q+1)
1
= 0 =
ΣΠ
(8)
II-o Pla+1)k-(q+1)t-q +1
A5g+6
h=0 k=1
t=0
Similarly
lim N-(9)
q+1)t-
t=0
lim N(7q+7)n+6q+7
2-(q+1)t-q
+1
n00
7h+6
1
ΣΠ
TI-, 2-(q+1)k-(q+1)t-q
+
h=0 k=1
=0
II, 2-(q+1)t-q/52_-(@) II
7h+6
1
A6q+7 =
2-(9) ( 1
II-o 2-(a+1)t-q +1
11-0 2-(4+1)k-(q+1)t-q † 1
t=0
h=0 k=1
3D0
o 7h+6
II-o 2-(q+1)t-q +1
II-, 2-(a+1)t-q/2-(a)
1
It=0
A6q+7
-ΣΠ
(9)
II-o P(a+1)k-(4+1)t-q +1
t30
h=0 k=1
=0
From the equations (3) and (4),
7h
II-o 2-(a+1)t-q+1
ΣΠ
ZII É lat+1)k-(4+1)t-q ™ -
1
>
=D0
-(q+1)t-q
h=0 k=1
t=0
0o 7h+1
II-, 2-(4+1)t-q+1
II-, 2-(a+1)t-q/
1
ΣΠ
%3D0
Ω
-(6q+5)
II-o P(a+1)k-(q+1)t-q + 1
=0
h=0 k=1
=0
thus 2-(7g+6) > S2-(6q+5)·
From the equations (4) and (5),
0o 7h+1
II-o 2-(a+1)t–q +1
IT-o 2-(q+1)t-q/52-(6q+5)
1
ΣΠ
t=0
II-o P(a+1)k-(q+1)t-q
>
+ 1
h=0 k=1
0o 7h+2
II-, 2-(4+1)t-q+1
1
ΣΠ
t=0
-(q+1)t-q/52-(5q+4)
II-, Pla+1)k-(a+1)t-q +1
h=0 k=1
thus N-(6q+5) > N-(5q+4).
*From the equations (5) and (6),
0o 7h+2
II-o 2-(a+1)t-q+1
II-0 2-(q+1)t-q/2-(5q+4)
1
It=0
ΣΠ
>
II-o P(a+1)k-(q+1)t-q +1
h=0 k=1
9
Transcribed Image Text:In this work, we deal with the following nonlinear difference equation
m-(7g+6)
1+II, ?m-(a+1)t-g
where N-(7g+6), SN-(79+5), ..., N-1, Slo E (0, 00) is investigated.
Sim+1
m 0, 1,..,
(1)
(e) If (7g+7)n+(t–1)q+t → a(t-1)q+t +0 then 2(7g+7)n+6q+7 +0 as n → o,
If 2(79+7)n+tg+t → arg+t +0 then 2(7g+7)n+7g+7 + 0 as n → 0o. t = 1,6.
(d) We can generate the following formulas:
(79+7)n+r(q+1)+s+1
(r-7)(a+1)+s+1 (1–
-(mq+m-1)+s)/S(r-7)(q+1)+s+1
1+ (IIm=1
2-(mg+m-1)+s)
7h+r
ΣΠ
IIm=1 P(a+1)t-(mq+m-1)+s + 1
h=0 k=1
lim 7a+7)n+g+2
= lim 2-(6q+5)
1
II-on-(9+1)t-g+1
n-00
n 7h+1
1
ΣΠ
IT-o 2-(a+1)k-(a+1)t-q +1,
h=0 k=1
Ag+2 = 2-(6q+5)
II-, 2-(9+1)t-g +1
00 7h+1
1
ΣΠ
II-o 2-(9+1)k-(q+1)t-q +1
h=0 k=1
00 7h+1
II-o 2-(9+1)t-9 +1
II, n-(a+1)t-g/2-(69+5)
1
ΣΠ
(4)
IIL, Pla+1)k-(a+1)t-g +1
ag+2 = 0 =
h=0 k=1
lim 7g+7)n+2q+3
= lim 2-(5g+4)
IT-o 2-(9+1)t-g+1
n400
n 7h+2
1
ΣΠ
II-o 2-(9+1)k-(a+1)t-q +
h=0 k=1
a2g+3 =
1-(54+4)
II-, 2-(a+1)t-g+1
n 7h+2
ΣΠ
II-, 2-(a+1)k-(a+1)t-q
+1
h=0 k=1
00 7h+2
II-, 2-(a+1)t-q +1
II,N-(a+1)t-a/2-(5a+4)
1
-ΣΠ
(5)
II, Pla+1)k-(a+1)t-g +1
aza+3
= 0 =
h=0 k=1
...
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