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 periodtwo, Р, Q, Р, Q, .of Eq.(1.1). From Eq.(1.1) we havea1Ym-1+ a2Ym–2+ a3Ym-3 + a4Ym-4+a5Ym-5B1ym-1+ B2ym-2 + B3Ym-3 + B4Ym-4 + B5Ym–5Ym+1 =Аут +(a1 + a3 + a5)Q+ (a2 + a4) P(B1 + B3 + B5) Q + (B2 + B4) P '(4.14)(a1 + a3 + a5) P+(a2+a4) QP = AQ+Q = AP+(B1 + B3 + B5) P+ (82 + B4) Q ³Consequently, we get(31 + Вз + Bs) Р? + (8ә + Ba) PQА (B + Вз + B5) РQ + A(3> + BA)Q+ (а1 + аз + а5) Р + (а2 + a4) Q,(4.15)andA (B1 + 3 + B5) PQ + A (B2 + B4) P²+ (a1 + a3 + a5) Q+ (a2 + a4) P.(4.16)(B1 + 33 + B5) Q² + (82 + B4) PQBy 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 ofthe solutions of the nonlinear difference equationa1Ym-1+a2Ym-2+ a3Ym-3+ a4Ym–4+ a5Ym-5Ут+1 — Аутtт 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 thatthe 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 caseB4when a4 =B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in thespecial case when az = B5 = 0.

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

Answered: Theorem 6 If (a1 + a3 + a5) > (a2 + a4)…

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

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

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

$y_{m + 1} = A y_{m} + \frac{α_{1} y_{m - 1} + α_{2} y_{m - 2} + α_{3} y_{m - 3} + α_{4} y_{m - 4} + α_{5} y_{m - 5}}{β_{1} y_{m - 1} + β_{2} y_{m - 2} + β_{3} y_{m - 3} + β_{4} y_{m - 4} + β_{5} y_{m - 5}}$ (1)

Also given that

$(α_{1} + α_{3} + α_{5}) > (α_{2} + α_{4}) a n d (β_{1} + β_{3} + β_{5}) > (β_{2} + β_{4})$

Since positive solutions are of prime period.

Let Solutions are $P, Q, P, Q, . . .$

If $y_{m + 1} = P, t h e n y_{m} = Q, y_{m - 1} = P, . . .$

and if

$y_{m + 1} = Q, t h e n y_{m} = P, y_{m - 1} = Q, . . .$

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