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 + B5) – (B2 + B4))] [(a1 + a3 + a5) – (a2 + a4)]? +4 [(a1 + a3 +a5) – (a2 + a4)] [(ß1 + B3 + B3) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] > 0. (4.13) is valid.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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How to deduce this equation from Equation 1.1 Explain to me the method. Show me the steps of determine blue and equation 1.1 is second pic

Theorem 6 If (a1 + a3 + a5) > (a2 + a4) and (ß1 + 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) ((81 + B3 + B5) – (32 + B4))] [(a1+ az + a5) – (a2 + a4)]?
+4 [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] > 0.
(4.13)
is valid.
proof: Suppose there exist positive distinctive solutions of prime period
two
, Р, Q, Р, Q, .
of Eq.(1.1). From Eq.(1.1) we have
a1Ym-1+ a2Ym–2+ a3Ym-3 + a4Ym-4+a5Ym-5
B1ym-1+ B2ym-2 + B3Ym-3 + B4Ym-4 + B5Ym–5
Ym+1 =
Аут +
(a1 + a3 + a5)Q+ (a2 + a4) P
(B1 + B3 + B5) Q + (B2 + B4) P '
(4.14)
(a1 + a3 + a5) P+(a2+a4) Q
P = AQ+
Q = AP+
(B1 + B3 + B5) P+ (82 + B4) Q ³
Consequently, we get
(31 + Вз + Bs) Р? + (8ә + Ba) PQ
А (B + Вз + B5) РQ + A(3> + BA)Q
+ (а1 + аз + а5) Р + (а2 + a4) Q,
(4.15)
and
A (B1 + 3 + B5) PQ + A (B2 + B4) P²
+ (a1 + a3 + a5) Q+ (a2 + a4) P.
(4.16)
(B1 + 33 + B5) Q² + (82 + B4) PQ
By subtracting (4.15) from (4.16), we obtain
[(B1 + B3 + Bs) + A (B2 + B4)] (P² – Q²) = [(a1 + a3 + az) – (a2 + a4)] (P – Q).
11
Transcribed Image Text:Theorem 6 If (a1 + a3 + a5) > (a2 + a4) and (ß1 + 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) ((81 + B3 + B5) – (32 + B4))] [(a1+ az + a5) – (a2 + a4)]? +4 [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] > 0. (4.13) is valid. proof: Suppose there exist positive distinctive solutions of prime period two , Р, Q, Р, Q, . of Eq.(1.1). From Eq.(1.1) we have a1Ym-1+ a2Ym–2+ a3Ym-3 + a4Ym-4+a5Ym-5 B1ym-1+ B2ym-2 + B3Ym-3 + B4Ym-4 + B5Ym–5 Ym+1 = Аут + (a1 + a3 + a5)Q+ (a2 + a4) P (B1 + B3 + B5) Q + (B2 + B4) P ' (4.14) (a1 + a3 + a5) P+(a2+a4) Q P = AQ+ Q = AP+ (B1 + B3 + B5) P+ (82 + B4) Q ³ Consequently, we get (31 + Вз + Bs) Р? + (8ә + Ba) PQ А (B + Вз + B5) РQ + A(3> + BA)Q + (а1 + аз + а5) Р + (а2 + a4) Q, (4.15) and A (B1 + 3 + B5) PQ + A (B2 + B4) P² + (a1 + a3 + a5) Q+ (a2 + a4) P. (4.16) (B1 + 33 + B5) Q² + (82 + B4) PQ By subtracting (4.15) from (4.16), we obtain [(B1 + B3 + Bs) + A (B2 + B4)] (P² – Q²) = [(a1 + a3 + az) – (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.
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.
Expert Solution
Step 1

Given that

ym+1=Aym+α1ym-1+α2ym-2+α3ym-3+α4ym-4+α5ym-5β1ym-1+β2ym-2+β3ym-3+β4ym-4+β5ym-5                    (1)

Also given that

(α1+α3+α5)>(α2+α4) and (β1+β3+β5)>(β2+β4)

Since positive solutions are of prime period.

Let Solutions are P, Q, P, Q, ...

If ym+1=P, then ym=Q, ym-1=P, ...

and if

ym+1=Q, then ym=P, ym-1=Q, ...

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