The main focus of this article is to discuss some qualitative behavior of the solutions of the nonlinear difference equation a1Ym-1+&2Ym-2+ a3Ym-3+a4Ym-4+ a5Ym-5 Aym+ B1Ym-1+ B2ym-2+ B3Ym-3 + B4Ym-4 + B5ym-5 Ym+1 = m = 0, 1,2, .., (1.1) where the coefficients A, a;i, Bi E (0, 00), i = 1, ..., 5, while the initial condi- tions y-5,y-4,Y-3,y-2,y-1, yo are arbitrary positive real numbers. Note that the special case of Eq.(1.1) has been discussed in [4] when az = B3 = a4 = B5 = 0 and Eq.(1.1) has been studied in [8] in the special case B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the B4 when a4 = = a5 = special case when as = B5 = 0. %3D
The main focus of this article is to discuss some qualitative behavior of the solutions of the nonlinear difference equation a1Ym-1+&2Ym-2+ a3Ym-3+a4Ym-4+ a5Ym-5 Aym+ B1Ym-1+ B2ym-2+ B3Ym-3 + B4Ym-4 + B5ym-5 Ym+1 = m = 0, 1,2, .., (1.1) where the coefficients A, a;i, Bi E (0, 00), i = 1, ..., 5, while the initial condi- tions y-5,y-4,Y-3,y-2,y-1, yo are arbitrary positive real numbers. Note that the special case of Eq.(1.1) has been discussed in [4] when az = B3 = a4 = B5 = 0 and Eq.(1.1) has been studied in [8] in the special case B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the B4 when a4 = = a5 = special case when as = B5 = 0. %3D
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 green and inf is here
![(a1 + a3 + a5) P+(a2+a4) Q
Ул — Р 3D AQ +
- P
(B1 + B3 + B5) P + (B2 + B4) Q
[A (B1 + B3 + B6) – (B2 + B4)]PQ + A (B2 + B4) Q² – (B1 + Ba3 + Bs) P²
(B1 + B3 + B3) P + (B2 + B4) Q
(a1 + a3 + a5) P + (@2+a4)Q
(В + Вз + В5) Р + (В2 + Ba) Q
[A(B1+B3+Bs)-(82+B4)][S1][(B1+B3+Bs)(a2+a4)+A(B2+B4) (a1+a3+a5)]
[(B1+B3+Bs)+A(B2+B4)]*[((B2+B4)-(B1+B3+B5))(A+1)]
([(a1+a3+a5)-(a2+a4)]+8
(B1 + B3 + B5) (2(31+B3+Bs)+A(B2+Ba)]
+ (B2 + B4) (Citag+a5)-(a9+as)]-6
2((B1+B3+B5)+A(B2+B4)]
2
2
A (B2 + B4) (
(@i+a3+as)-(az+as)]-6
2[(B1+B3+B5)+A(B2+B4)]
-(81 + B3 + Bs)
[(@1+a3+as)-(a2+as)]+8
2(B1+B3+Bs)+A(ß2+B4)]
PI T P3 + P5) (231+33+85)+A(62+8) + (B2 + B4) (eIta3+as)-(a2+a4)]-8
[(ai+a3+as)-(a2+a4)]-8
2[(B1+B3+B5)+A(B82+B4)]
I(a1+a3+as)-(a2+a4)]-8
2((B1+33+B5)+A(B2+B4)
(4.24)
[(ai+a3+a5)-(a2+a4)]+8
2[(B1+B3+Bs)+A(B2+B4)]
(a1+a3+as)-(a2+a4)]+ô
2[(B1+B3+B5)+A(B2+B4)]
(a1+a3+as)-(aztas)]+8
2[(B1+B3+B3)+A(B2+B4)]
(a1 + a3 + a5)
+ (a2 + a4)
(B1 + B3 + B5)
+ (B2 + B4)
The main focus of this article is to discuss some qualitative behavior of
the solutions of the nonlinear difference equation
a1Ym-1+a2Ym-2 + a3Ym-3 + a4Ym-4 + a5Ym-5
Ym+1 =
Aym+
m = 0,1,2, ..,
В1ут-1 + В2ут-2 + Взут-3 + Ваут-4 + Bзут-5
(1.1)
where the coefficients A, a;, Bi E (0, 00), i = 1, .., 5, while the initial condi-
tions y-5,y-4,y-3,y-2,y-1, yo are arbitrary positive real numbers. Note that
the special case of Eq.(1.1) has been discussed in [4] when az =
B4
when a4 =
= a4 =
= 0 and Eq.(1.1) has been studied in [8] in the special case
B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the
= a5 =
= B5
special case when as = B5 = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2F0546e762-a861-49c2-8b9e-e71c18829446%2Fc8ckgyf_processed.png&w=3840&q=75)
Transcribed Image Text:(a1 + a3 + a5) P+(a2+a4) Q
Ул — Р 3D AQ +
- P
(B1 + B3 + B5) P + (B2 + B4) Q
[A (B1 + B3 + B6) – (B2 + B4)]PQ + A (B2 + B4) Q² – (B1 + Ba3 + Bs) P²
(B1 + B3 + B3) P + (B2 + B4) Q
(a1 + a3 + a5) P + (@2+a4)Q
(В + Вз + В5) Р + (В2 + Ba) Q
[A(B1+B3+Bs)-(82+B4)][S1][(B1+B3+Bs)(a2+a4)+A(B2+B4) (a1+a3+a5)]
[(B1+B3+Bs)+A(B2+B4)]*[((B2+B4)-(B1+B3+B5))(A+1)]
([(a1+a3+a5)-(a2+a4)]+8
(B1 + B3 + B5) (2(31+B3+Bs)+A(B2+Ba)]
+ (B2 + B4) (Citag+a5)-(a9+as)]-6
2((B1+B3+B5)+A(B2+B4)]
2
2
A (B2 + B4) (
(@i+a3+as)-(az+as)]-6
2[(B1+B3+B5)+A(B2+B4)]
-(81 + B3 + Bs)
[(@1+a3+as)-(a2+as)]+8
2(B1+B3+Bs)+A(ß2+B4)]
PI T P3 + P5) (231+33+85)+A(62+8) + (B2 + B4) (eIta3+as)-(a2+a4)]-8
[(ai+a3+as)-(a2+a4)]-8
2[(B1+B3+B5)+A(B82+B4)]
I(a1+a3+as)-(a2+a4)]-8
2((B1+33+B5)+A(B2+B4)
(4.24)
[(ai+a3+a5)-(a2+a4)]+8
2[(B1+B3+Bs)+A(B2+B4)]
(a1+a3+as)-(a2+a4)]+ô
2[(B1+B3+B5)+A(B2+B4)]
(a1+a3+as)-(aztas)]+8
2[(B1+B3+B3)+A(B2+B4)]
(a1 + a3 + a5)
+ (a2 + a4)
(B1 + B3 + B5)
+ (B2 + B4)
The main focus of this article is to discuss some qualitative behavior of
the solutions of the nonlinear difference equation
a1Ym-1+a2Ym-2 + a3Ym-3 + a4Ym-4 + a5Ym-5
Ym+1 =
Aym+
m = 0,1,2, ..,
В1ут-1 + В2ут-2 + Взут-3 + Ваут-4 + Bзут-5
(1.1)
where the coefficients A, a;, Bi E (0, 00), i = 1, .., 5, while the initial condi-
tions y-5,y-4,y-3,y-2,y-1, yo are arbitrary positive real numbers. Note that
the special case of Eq.(1.1) has been discussed in [4] when az =
B4
when a4 =
= a4 =
= 0 and Eq.(1.1) has been studied in [8] in the special case
B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the
= a5 =
= B5
special case when as = B5 = 0.
![Multiplying the denominator and numerator of (4.24) by 4 [(B1 + B3 + B5) + A (B2 + B4)]?
we get
4[A(B1+B3+B5)-(B2+B4)][S1][(B1+ß3+B5)(@2+a4)+A(B2+B4) (a1+a3+æ5)]
[(B2+B4)-(B1+B3+Bs))(A+1)]
Y1 – P =
A (B2 + B4) (S1 – 8)² – (B1 + B3 + Bs) (Sı + 8)²
S
2[(B1 + B3 + B5) + A (B2 + B4)] (aı + a3 + a5) (S1 + 8)
+
S
2[(B1 + B3 + B5) + A (B2 + B4)] (a2 + a4) (S1 – 8)
+
S
4[A(B1+B3+B5)-(B2+Ba)][S1][(B1+B3+Bs)(a2+a4)+A(B2+B4) (a1+a3+a5)]
[((32+B4)-(B1+ß3+B5))(A+1)]
A (B2 + Ba) [S1]? - (B1 + B3 + B5) [S1]?
[A (B2 + B4) – (ß1 + B3 + B3)]82
S
14
2[(B1 + B3 + B5) + A (B2 + B4)] (a2+a4) [S1]
S
2[(B1 + B3 + B5)+ A(B2 + B4)] (a1 + a3 + a5) [S1]
S
2A (B2 + B4) [(a1+az + a5) – (a2 + a4)] & + 2 (B1 + B3 + B5) [S1] &
S
2[(B1 + B3 + B5) + A (B2 + Ba)] (a1+ a3 + a5) 8 – 2[(B1 + B3 + B5) + A (B2 + Ba)] (a2 + a4) 8
S
4[A(B1+B3+B5)-(B2+B4)][S1][(B1+B3+B5)(a2+a4)+A(B2+B4)(a1+a3+as)]
[((B2+B4)-(B1+B3+B5))(A+1)]
S
4[Si][(B1+B3+ßs)(a2+as)+A(ß2+ß4)(@1+a3+as)][A(B2+Ba)-(B1+B3+ßs)1
[((B2+B4)-(B1+ß3+B5))(A+1)]
A (B2 + B4) [S1]? – (B1 + B3 + B3) [S1]²
-
S
[A (B2 + B4) – (ß1 + B3 + Bs)] [S1]²
S
2[(B1 + B3 + Bs) + A (82 + B4)] (a2+a4) [S1]
+
S
2[(B1 + B3 + B3)+ A (B2 + B4)] (a1 +a3 + as) [S1]
+
2 [S1] [(B1 + B3 + Bs) + A (B2 + B4)]8
S
2 [S1] [(B1 + B3 + Bs) + A (82 + B4)]S
+
S
4 [S1] [(B1 + B3 + Bg) (a2 + a4) + A (B2 + B4) (a1 + az + a5)]
S
4 [S1] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)]
S
2[(a1 + a3 + as) – (a2 + a4)][(B1 + B3 + B5) + A (ß2 + B4)]8
S
2[(a1 + az + a5) – (a2 + a4)][(B1 + B3 + B3) + A (B2 + B4)]8
S
= 0.
where
2 [(B1 + B3 + B5)+A (B2 + B4)] ×
[(B1 + B3 + B5) (S, + 6) + (B2 + B4) (S1 – 8)]
S
=
-
15](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2F0546e762-a861-49c2-8b9e-e71c18829446%2Fusqkolk_processed.png&w=3840&q=75)
Transcribed Image Text:Multiplying the denominator and numerator of (4.24) by 4 [(B1 + B3 + B5) + A (B2 + B4)]?
we get
4[A(B1+B3+B5)-(B2+B4)][S1][(B1+ß3+B5)(@2+a4)+A(B2+B4) (a1+a3+æ5)]
[(B2+B4)-(B1+B3+Bs))(A+1)]
Y1 – P =
A (B2 + B4) (S1 – 8)² – (B1 + B3 + Bs) (Sı + 8)²
S
2[(B1 + B3 + B5) + A (B2 + B4)] (aı + a3 + a5) (S1 + 8)
+
S
2[(B1 + B3 + B5) + A (B2 + B4)] (a2 + a4) (S1 – 8)
+
S
4[A(B1+B3+B5)-(B2+Ba)][S1][(B1+B3+Bs)(a2+a4)+A(B2+B4) (a1+a3+a5)]
[((32+B4)-(B1+ß3+B5))(A+1)]
A (B2 + Ba) [S1]? - (B1 + B3 + B5) [S1]?
[A (B2 + B4) – (ß1 + B3 + B3)]82
S
14
2[(B1 + B3 + B5) + A (B2 + B4)] (a2+a4) [S1]
S
2[(B1 + B3 + B5)+ A(B2 + B4)] (a1 + a3 + a5) [S1]
S
2A (B2 + B4) [(a1+az + a5) – (a2 + a4)] & + 2 (B1 + B3 + B5) [S1] &
S
2[(B1 + B3 + B5) + A (B2 + Ba)] (a1+ a3 + a5) 8 – 2[(B1 + B3 + B5) + A (B2 + Ba)] (a2 + a4) 8
S
4[A(B1+B3+B5)-(B2+B4)][S1][(B1+B3+B5)(a2+a4)+A(B2+B4)(a1+a3+as)]
[((B2+B4)-(B1+B3+B5))(A+1)]
S
4[Si][(B1+B3+ßs)(a2+as)+A(ß2+ß4)(@1+a3+as)][A(B2+Ba)-(B1+B3+ßs)1
[((B2+B4)-(B1+ß3+B5))(A+1)]
A (B2 + B4) [S1]? – (B1 + B3 + B3) [S1]²
-
S
[A (B2 + B4) – (ß1 + B3 + Bs)] [S1]²
S
2[(B1 + B3 + Bs) + A (82 + B4)] (a2+a4) [S1]
+
S
2[(B1 + B3 + B3)+ A (B2 + B4)] (a1 +a3 + as) [S1]
+
2 [S1] [(B1 + B3 + Bs) + A (B2 + B4)]8
S
2 [S1] [(B1 + B3 + Bs) + A (82 + B4)]S
+
S
4 [S1] [(B1 + B3 + Bg) (a2 + a4) + A (B2 + B4) (a1 + az + a5)]
S
4 [S1] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)]
S
2[(a1 + a3 + as) – (a2 + a4)][(B1 + B3 + B5) + A (ß2 + B4)]8
S
2[(a1 + az + a5) – (a2 + a4)][(B1 + B3 + B3) + A (B2 + B4)]8
S
= 0.
where
2 [(B1 + B3 + B5)+A (B2 + B4)] ×
[(B1 + B3 + B5) (S, + 6) + (B2 + B4) (S1 – 8)]
S
=
-
15
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
Step 1
The equation given to us is as follows:
When the RHS is simplified we will get the equation as follows:
Here .
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