Show me the steps of determine yellow and the informatian is here 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. 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)] [(a1 + a3 + as) – (a2 + a4)]?+4 [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + az + as)] > 0.(4.13)is valid.proof: Suppose there exist positive distinctive solutions of prime periodtwo..., P, Q, P,Q, .....of Eq.(1.1). From Eq.(1.1) we havea1Ym–1+a2Ym–2+a3Ym-3+¤4Ym-4+a5Ym–5Ym+1 = Aym+Biym-1 + B2ym-2 + B3Ym-3 + B4Ym-4 + B5Ym-5(a1 + a3 + a5)P+ (a2 + a4) Q(B1 + B3 + Bs) P+(B2 + B4) Q(a1 + a3 + a5) Q+ (a2 + a4) P(B1 + B3 + Bs) Q+ (B2 + B4) P(4.14)P = AQ+Q = AP+Consequently, we get(B1 + B3 + Bs) P² + (B2 + B4) PQ = A(B1+ B3 + B5) PQ + A (B2 + B4) Q?+(a1+ a3 + a5) P+ (a2+a4) Q,(4.15)and(B1 + B3 + B3) Q² + (B2 + B4) PQ = A (B1 + B3 + B5) PQ + A (B2 + B4) p²+ (a1 + a3 + a5) Q+ (a2 + a4) P.(4.16)By subtracting (4.15) from (4.16), we obtain[(B1 + B3 + Bs) + A (B2 + B4)] (P² – Q²) = [(a1 + a3 +a5) – (a2 + a4)] (P – Q).11Since P+ Q, it follows that[(a1 + a3 + a5) – (a2 + a4)][(B1 + B3 + B5) + A (B2 + B4)]'P+Q =(4.17)while, by adding (4.15) and (4.16) and by using the relationp² + Q? = (P+ Q)² – 2PQ for allP,Q ER,we have[(a1 + az + a5) - (a2 + a4)] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + az + a5)][(B1 + B3 + B5) + A (B2 + B4)]* [((B2 + B4) – (B1 + B3 + B5)) (A+1)]PQ =(4.18)Let P and Q are two distinct real roots of the quadratic equationt2 - (P+ Q)t + PQ = 0.[(B1 + B3 + B5) + A (B2 + B4)] t – [(@1 + a3 + a5) – (a2 + a4)]t[(a1 + az + a5) - (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)]+[(81 + B3 + B5) + A (B2 + B4)] [((B2 + B4) – (B1 + B3 + Bs)) (A+1)]0,(4.19)

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Description: Show me the steps of determine yellow and the informatian is here 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,

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[(@1 + a3 + a5) - (a2 + a4)] [(B1 + B3 + B3) + A (B2 + B4)]' P+Q = (4.17) %3D while, by adding (4.15) and (4.16) and by using the relation p² + Q? = (P+Q)² – 2PQ for al P,Q E R, we have [(a1 + a3 + a5) - (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] PQ [(B1 + B3 + B3) + A (B2 + B4)]² [((B2 + B4) – (B1 + B3 + B5)) (A+1)] (4.18) Let P and Q are two distinct real roots of the quadratic equation

[(@1 + a3 + a5) - (a2 + a4)] [(B1 + B3 + B3) + A (B2 + B4)]' P+Q = (4.17) %3D while, by adding (4.15) and (4.16) and by using the relation p² + Q? = (P+Q)² – 2PQ for al P,Q E R, we have [(a1 + a3 + a5) - (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] PQ [(B1 + B3 + B3) + A (B2 + B4)]² [((B2 + B4) – (B1 + B3 + B5)) (A+1)] (4.18) Let P and Q are two distinct real roots of the quadratic equation

Advanced Engineering Mathematics

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ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Show me the steps of determine yellow and the informatian is here

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.

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)] [(a1 + a3 + as) – (a2 + a4)]?
+4 [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + az + as)] > 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
a1Ym–1+a2Ym–2+a3Ym-3+¤4Ym-4+a5Ym–5
Ym+1 = Aym+
Biym-1 + B2ym-2 + B3Ym-3 + B4Ym-4 + B5Ym-5
(a1 + a3 + a5)P+ (a2 + a4) Q
(B1 + B3 + Bs) P+(B2 + B4) Q
(a1 + a3 + a5) Q+ (a2 + a4) P
(B1 + B3 + Bs) Q+ (B2 + B4) P
(4.14)
P = AQ+
Q = AP+
Consequently, we get
(B1 + B3 + Bs) P² + (B2 + B4) PQ = A(B1+ B3 + B5) PQ + A (B2 + B4) Q?
+(a1+ a3 + a5) P+ (a2+a4) Q,
(4.15)
and
(B1 + B3 + B3) Q² + (B2 + B4) PQ = A (B1 + B3 + B5) PQ + A (B2 + B4) p²
+ (a1 + a3 + a5) Q+ (a2 + a4) P.
(4.16)
By subtracting (4.15) from (4.16), we obtain
[(B1 + B3 + Bs) + A (B2 + B4)] (P² – Q²) = [(a1 + a3 +a5) – (a2 + a4)] (P – Q).
11
Since P+ Q, it follows that
[(a1 + a3 + a5) – (a2 + a4)]
[(B1 + B3 + B5) + A (B2 + B4)]'
P+Q =
(4.17)
while, by adding (4.15) and (4.16) and by using the relation
p² + Q? = (P+ Q)² – 2PQ for all
P,Q ER,
we have
[(a1 + az + a5) - (a2 + a4)] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + az + a5)]
[(B1 + B3 + B5) + A (B2 + B4)]* [((B2 + B4) – (B1 + B3 + B5)) (A+1)]
PQ =
(4.18)
Let P and Q are two distinct real roots of the quadratic equation
t2 - (P+ Q)t + PQ = 0.
[(B1 + B3 + B5) + A (B2 + B4)] t – [(@1 + a3 + a5) – (a2 + a4)]t
[(a1 + az + a5) - (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)]
+
[(81 + B3 + B5) + A (B2 + B4)] [((B2 + B4) – (B1 + B3 + Bs)) (A+1)]
0,
(4.19)

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It is given that, the value of P and Q is as follows:

...... (1)

and

......(2)

To prove:

we have to prove that,

and

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