Explain the determine blue and the inf is here Theorem 6 If (¤1 + 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 + B3) – (B2 + B4))[@ + a3 + as) – (a2 + a4)]?+4[(a1 + a3 + a5) – (a2 + a4)] [(ß1 + B3 + Bs) (a2 +a4) + A (B2 + B4) (a1 + a3 + a5)] > 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 + a4Ym-4 +a5Ym-5B1ym-1 + B2ym-2 + B3Ym-3 + B4Ym-4 + B5Ym-5Ym+1 =Aym +(a1 + a3 + a5) P+(a2+a4) Q(B1 + B3 + B5) P+ (B2 + B4) Q'(ai + a3 + a5)Q+(@2+a4) P(B1 + B3 + B5) Q + (B2 + B4) P(4.14)P = AQ+Q— АР+Consequently, we get(B1 + B3 + B5) P2 + (82 + B4) PQ = A(B1 +B3 + B5) PQ + A (B2 + B4) Q²+ (a1 + a3 + a5) P+ (a2 + a4) Q,(4.15)and(B1 + B3 + Bs) Q? + (B2 + Ba) PQA (B1 + B3 + Bs) PQ + A (B2 + B4) P²+ (a1 + a3 + a5) Q + (a2 + a4) P.(4.16)By subtracting (4.15) from (4.16), we obtain[(31 + Ba + Bs) + A (82 + B4)] (P² – Q²) = [(a1 + a3 + as) – (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 – 2PQfor allP,Q E R,we have[(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)][(B1 + B3 + B5) + A (82 + B4)]? [((B2 + B4) – (B1 + B3 + B3)) (A + 1)](4.18)PQ =Let P and Q are two distinct real roots of the quadratic equationt2 - (P+Q)t+ PQ = 0.[(B1 + B3 + B5) + A (B2 + Ba)] t² – [(a1 + a3 + a5) – (a2 + a4)] t[(a1 + a3 + as) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)][(B1 + B3 + B5) + A (B2 + B4)] [((B2 + B4) – (B1 + B3 + B5)) (A + 1)]-0,(4.19)and soC [(a1 + a3 + a5) – (a2 +a4)]?(a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + az + a5)][(B2 + B4) – (B1 + B3 + Bs)) (A+1)]4 [(a1 + a3 +a5) –> 0,or[(a1 + a3 + as) – (a2 + a4)]?(a2 + a4)] [(B1 + ß3 + B5) (a2 + a4) + A (B2 + B4) (a+a3 + a5)][((P1 + B3 + B3) – (82 + B4)) (A + 1)]4 [(a1 + a3 + 5) –> 0.(4.20)From (4.20), we get[(В1 + Вз + Bs) - (32 + Bа)) (А + 1)] [(а1 + аз + as) — (а2 + aд)}?+4 [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] > 0.Therefore, the condition (4.13) is valid. Alternatively, if we imagine that thecondition (4.13) is valid where (a1+a3 + a5) > (a2 + a4) and (B1 + B3 + B5) >(B2 + B4). Then, we can immediately discover that the inequality stands. 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-5Ym+1 =Aym+т 3D 0, 1, 2, ...,В1ут-1 + В2ут-2 + Взут-3 + Влут-4 + B5ут-5(1.1)

Answered: Image /qna-images/answer/bcf90ef4-c785-46b6-a5bb-ebf06590aa18.jpg

[(B1 + B3 + B5) + A (B2 + B4)] t² – [(a1 + a3 + a5) – (a2+ a4)]t [(ат + аз + а5) —- (аэ + ад)] [(31 + Вз + B5) (аэ + a4) + A (B2 + Bа) (а + аз + as)] + [(B1 + B3 + Bs) + A (B2 + B4)] [((B2 + B4) – (B1 + B3 + B5)) (A + 1)] - = 0, (4.19) and so C [(a1 + a3 + a5) – (a2 + a4)]² 4 [(a1 + a3 + a5) – (a2 + a4)] [(ß1 + B3 + B5) (a2+a4) + A (B2 + B4) (a1 + a3 + a5)] [((B2 + B4) – (B1 + B3 + Bs)) (A + 1)I > 0, or [(a1 + a3 + as) – (a2 + a4)]? 4 [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (@1 + a3 + a5)] [((B1 + B3 + B5) - (B2 + B4)) (A + 1)] > 0. (4.20) From (4.20), we get

[(B1 + B3 + B5) + A (B2 + B4)] t² – [(a1 + a3 + a5) – (a2+ a4)]t [(ат + аз + а5) —- (аэ + ад)] [(31 + Вз + B5) (аэ + a4) + A (B2 + Bа) (а + аз + as)] + [(B1 + B3 + Bs) + A (B2 + B4)] [((B2 + B4) – (B1 + B3 + B5)) (A + 1)] - = 0, (4.19) and so C [(a1 + a3 + a5) – (a2 + a4)]² 4 [(a1 + a3 + a5) – (a2 + a4)] [(ß1 + B3 + B5) (a2+a4) + A (B2 + B4) (a1 + a3 + a5)] [((B2 + B4) – (B1 + B3 + Bs)) (A + 1)I > 0, or [(a1 + a3 + as) – (a2 + a4)]? 4 [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (@1 + a3 + a5)] [((B1 + B3 + B5) - (B2 + B4)) (A + 1)] > 0. (4.20) From (4.20), we get

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

Explain the determine blue and the inf is here

Theorem 6 If (¤1 + 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 + B3) – (B2 + B4))[@ + a3 + as) – (a2 + a4)]?
+4[(a1 + a3 + a5) – (a2 + a4)] [(ß1 + B3 + Bs) (a2 +a4) + A (B2 + B4) (a1 + a3 + a5)] > 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 + a4Ym-4 +a5Ym-5
B1ym-1 + B2ym-2 + B3Ym-3 + B4Ym-4 + B5Ym-5
Ym+1 =
Aym +
(a1 + a3 + a5) P+(a2+a4) Q
(B1 + B3 + B5) P+ (B2 + B4) Q'
(ai + a3 + a5)Q+(@2+a4) P
(B1 + B3 + B5) Q + (B2 + B4) P
(4.14)
P = AQ+
Q— АР+
Consequently, we get
(B1 + B3 + B5) P2 + (82 + B4) PQ = A(B1 +B3 + B5) PQ + A (B2 + B4) Q²
+ (a1 + a3 + a5) P+ (a2 + a4) Q,
(4.15)
and
(B1 + B3 + Bs) Q? + (B2 + Ba) PQ
A (B1 + B3 + Bs) PQ + A (B2 + B4) P²
+ (a1 + a3 + a5) Q + (a2 + a4) P.
(4.16)
By subtracting (4.15) from (4.16), we obtain
[(31 + Ba + Bs) + A (82 + B4)] (P² – Q²) = [(a1 + a3 + as) – (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 E R,
we have
[(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)]
[(B1 + B3 + B5) + A (82 + B4)]? [((B2 + B4) – (B1 + B3 + B3)) (A + 1)]
(4.18)
PQ =
Let P and Q are two distinct real roots of the quadratic equation
t2 - (P+Q)t+ PQ = 0.
[(B1 + B3 + B5) + A (B2 + Ba)] t² – [(a1 + a3 + a5) – (a2 + a4)] t
[(a1 + a3 + as) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)]
[(B1 + B3 + B5) + A (B2 + B4)] [((B2 + B4) – (B1 + B3 + B5)) (A + 1)]
-
0,
(4.19)
and so
C [(a1 + a3 + a5) – (a2 +a4)]?
(a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + az + a5)]
[(B2 + B4) – (B1 + B3 + Bs)) (A+1)]
4 [(a1 + a3 +a5) –
> 0,
or
[(a1 + a3 + as) – (a2 + a4)]?
(a2 + a4)] [(B1 + ß3 + B5) (a2 + a4) + A (B2 + B4) (a+a3 + a5)]
[((P1 + B3 + B3) – (82 + B4)) (A + 1)]
4 [(a1 + a3 + 5) –
> 0.
(4.20)
From (4.20), we get
[(В1 + Вз + Bs) - (32 + Bа)) (А + 1)] [(а1 + аз + as) — (а2 + aд)}?
+4 [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] > 0.
Therefore, the condition (4.13) is valid. Alternatively, if we imagine that the
condition (4.13) is valid where (a1+a3 + a5) > (a2 + a4) and (B1 + B3 + B5) >
(B2 + B4). Then, we can immediately discover that the inequality stands.

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+
т 3D 0, 1, 2, ...,
В1ут-1 + В2ут-2 + Взут-3 + Влут-4 + B5ут-5
(1.1)

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