(C (g+1)n-((q+1)²-1) (g+1)n-[(g+2)²+(5q+3)] n-(q+1) (₁-2-(79+6)) II 1 + [²_o (q+1)k-(q+1)t-q k=1 = ⠀ 1 (g+1)n-((a+1)²-(g+1)) — (g+1)n-[(q+2)²+(4q+3)]' Ω n-(q+1) = (₁+1 -2-(6q+6)) II k=1 1 1+ i=0 (g+1)k-(q+1)t
(C (g+1)n-((q+1)²-1) (g+1)n-[(g+2)²+(5q+3)] n-(q+1) (₁-2-(79+6)) II 1 + [²_o (q+1)k-(q+1)t-q k=1 = ⠀ 1 (g+1)n-((a+1)²-(g+1)) — (g+1)n-[(q+2)²+(4q+3)]' Ω n-(q+1) = (₁+1 -2-(6q+6)) II k=1 1 1+ i=0 (g+1)k-(q+1)t
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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Transcribed Image Text:In this work, we deal with the following nonlinear difference equation
Im-(79+6)
1+ II-o Pm-(9+1)t-g
Im+1
= 0, 1, ..,
(1)
m =
t%3D0
where N-(79+6), 2-(7g+5), ·
-1, No E (0, 0) is investigated.
(d) We can generate the following formulas:
L(79+7)n+r(q+1)+s+1
r-7)(4+1)+s+1 (1–
(ITm-12-(mq+,
1+ (IIm=1 2-
m-1)+s)/2(r-7)(a+1)+s+1
(mq+m-:
n-1)+s)
Lm3D
n
7h+r
ΣΠ
IIm=1 P(a+1)t-(mq+m-1)+s + 1
1
h=0 k=1
(e) If 2(7g+7)n+(t–1)q+t → a(t-1)q+t 70 then 2(7g+7)n+6q+7 → 0 as n → o,
If 2(79+7)n+tq+t → atg+t #0 then 2(7g+7)n+7q+7 → 0 as n → o. t = 1,6.
![(d) Subtracting Nm-(7g+6) from both sides in (1), we have:
1
(Pm
1+II-, Pm-(q+1)t-q
Im+1
Im-(7q+6)
?m-q – Sm-(8q+7),
t=0
and the following formula, for n > q+1
P(a+1)n-(9+1)²–1)¯
Pa+1)n-[(a+2)²+(5q+3)]
п- (q+1)
1
= (L1 – N-(7g+6) II
1+II-o P(q+1)k-(a+1)t-q
k=1
t=0
P(a+1)n-(a+1)°-(a+1)) - Pa+1)n=[(q+2)*+(4q+3) :
n-(q+1)
1
= (Lq+1 – N-(6g+6) II
1+II?-o ?(a+1)k-(q+1)t
k=1
From (2) we get:
7h
1
L(79+7)n+1 - 2-(7q+6)
(2 - N-(7a+6)) 2IIÉLa+1)k-(q+1)t-q + 1
h=0 k=1
t=0
7h
1
L(Ta+7)a+q+1 - 2-(6g+6) = (Lg+1 – 2-(6g+6) II
h=0 k=111-0 (q+1)k-(q+1)t + 1
So
7h+1
1
(7g+7)n+g+2 - N-(6q+5) =
(2 - 2-(7q+6)) II
P(a+1)k-(a+1)t-q
+ 1'
h=0 k=1
4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F158aed3a-44ed-4147-bb6c-9729d6a7101b%2F6e552bb3-84fb-42f6-83d0-09252004d081%2Fvcc99mh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(d) Subtracting Nm-(7g+6) from both sides in (1), we have:
1
(Pm
1+II-, Pm-(q+1)t-q
Im+1
Im-(7q+6)
?m-q – Sm-(8q+7),
t=0
and the following formula, for n > q+1
P(a+1)n-(9+1)²–1)¯
Pa+1)n-[(a+2)²+(5q+3)]
п- (q+1)
1
= (L1 – N-(7g+6) II
1+II-o P(q+1)k-(a+1)t-q
k=1
t=0
P(a+1)n-(a+1)°-(a+1)) - Pa+1)n=[(q+2)*+(4q+3) :
n-(q+1)
1
= (Lq+1 – N-(6g+6) II
1+II?-o ?(a+1)k-(q+1)t
k=1
From (2) we get:
7h
1
L(79+7)n+1 - 2-(7q+6)
(2 - N-(7a+6)) 2IIÉLa+1)k-(q+1)t-q + 1
h=0 k=1
t=0
7h
1
L(Ta+7)a+q+1 - 2-(6g+6) = (Lg+1 – 2-(6g+6) II
h=0 k=111-0 (q+1)k-(q+1)t + 1
So
7h+1
1
(7g+7)n+g+2 - N-(6q+5) =
(2 - 2-(7q+6)) II
P(a+1)k-(a+1)t-q
+ 1'
h=0 k=1
4
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