utions of prime period two is that the inequality [(A+1) ((B1 + B3 + B5) – (B2 + B4)] [(a1 + a3 + az) – (a2 + a4)]? +4 [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] > 0. (4.13)
utions of prime period two is that the inequality [(A+1) ((B1 + B3 + B5) – (B2 + B4)] [(a1 + a3 + az) – (a2 + a4)]? +4 [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] > 0. (4.13)
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 the inf is here
![Theorem 6 If (a1 + a3 + a5) > (a2 + a4) and (B1 + B3 + B5) > (B2 + B4),
then the necessary and sufficient condition for Eq.(1.1) to have positive so-
lutions of prime period two is that the inequality
[(A+1) ((B1 + B3 + Bs) – (B2 + B4))] [(@1 + a3 + a5) – (a2 + a4)]?
+4 [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (@1 + az + az)] > 0.
(4.13)
is valid.
proof: Suppose there exist positive distinctive solutions of prime period
two
P,Q, P, Q,..
......
of Eq.(1.1). From Eq.(1.1) we have
а1Ут -1 + ӕ2Ут -2 + азут -3 + a4ym-4 + a5ym-5
Ym+1 = Aym +
B1Ym-1+ B2ym–2 + B3Ym-3 + B4Ym-4 + B5Ym-5
(а1 + оз + аs) Р + (02 + а4) Q
AQ+
(B1 + B3 + B5) P + (B2 + B4) Q ’
(a1 + a3 + a5) Q+ (a2 + a4) P
(B1 + B3 + B5) Q + (B2 + B4) P '
(4.14)
Q3 АР+
Consequently, we get
A (B1 + B3 + Bs) PQ + A (B2 + B4) Q²
+ (a1 + a3 + a5) P + (a2 + a4) Q,
(4.15)
(B1 + B3 + B5) P2 + (B2 + B4) PQ
and
(B1 + B3 + B5) Q² + (B2 + B4) PQ = A(ß1+ B3 + B5) PQ + A (B2 + B4) P²
+ (@1 + a3 + a5) Q + (a2 + a4) P.
(4.16)
By subtracting (4.15) from (4.16), we obtain
[(B1 + B3 + B5) + A (B2 + B4)] (P² – Q²) = [(a1 + a3 + a5) – (a2 + a4)] (P – Q).
11](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2F2ec5d1d9-642c-40fa-9ffc-401cea4c9a32%2F5xw1vb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 6 If (a1 + a3 + a5) > (a2 + a4) and (B1 + B3 + B5) > (B2 + B4),
then the necessary and sufficient condition for Eq.(1.1) to have positive so-
lutions of prime period two is that the inequality
[(A+1) ((B1 + B3 + Bs) – (B2 + B4))] [(@1 + a3 + a5) – (a2 + a4)]?
+4 [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (@1 + az + az)] > 0.
(4.13)
is valid.
proof: Suppose there exist positive distinctive solutions of prime period
two
P,Q, P, Q,..
......
of Eq.(1.1). From Eq.(1.1) we have
а1Ут -1 + ӕ2Ут -2 + азут -3 + a4ym-4 + a5ym-5
Ym+1 = Aym +
B1Ym-1+ B2ym–2 + B3Ym-3 + B4Ym-4 + B5Ym-5
(а1 + оз + аs) Р + (02 + а4) Q
AQ+
(B1 + B3 + B5) P + (B2 + B4) Q ’
(a1 + a3 + a5) Q+ (a2 + a4) P
(B1 + B3 + B5) Q + (B2 + B4) P '
(4.14)
Q3 АР+
Consequently, we get
A (B1 + B3 + Bs) PQ + A (B2 + B4) Q²
+ (a1 + a3 + a5) P + (a2 + a4) Q,
(4.15)
(B1 + B3 + B5) P2 + (B2 + B4) PQ
and
(B1 + B3 + B5) Q² + (B2 + B4) PQ = A(ß1+ B3 + B5) PQ + A (B2 + B4) P²
+ (@1 + a3 + a5) Q + (a2 + a4) P.
(4.16)
By subtracting (4.15) from (4.16), we obtain
[(B1 + B3 + B5) + A (B2 + B4)] (P² – Q²) = [(a1 + a3 + a5) – (a2 + a4)] (P – Q).
11
![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
Ут+1 — Аутt
т 3 0, 1, 2, ...,
B1Ym-1+ B2Ym-2 + B3Ym-3 + B4Ym-4 + B3Ym-5
(1.1)
where the coefficients A, a;, B; E (0, 0), 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 =
= a5 = B5 = 0 and Eq.(1.1) has been studied in [8] in the special case
B4
when a4 =
B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the
special case when az = B5 = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2F2ec5d1d9-642c-40fa-9ffc-401cea4c9a32%2Fd1y9kjx_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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
Ут+1 — Аутt
т 3 0, 1, 2, ...,
B1Ym-1+ B2Ym-2 + B3Ym-3 + B4Ym-4 + B3Ym-5
(1.1)
where the coefficients A, a;, B; E (0, 0), 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 =
= a5 = B5 = 0 and Eq.(1.1) has been studied in [8] in the special case
B4
when a4 =
B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the
special case when az = B5 = 0.
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