Since P+ Q, it follows that [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) + A (82 + 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 all P,Q E R, we have [(a1 + a3 + a5) - (a2+ a4)] [(B1 + B3 + B3) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] [(B1 + B3 + B5) + A (B2 + B4)] [((B2 + Ba) – (B1 + B3 + B3)) (A+1)] PQ (4.18) Let P and Q are two distinct real roots of the quadratic equation t - (P+Q)t+ PQ = 0. %3D [(B1 + B3 + B5) + A (B2 + B4)] t² – [(a1 + a3 + a5) –- (a2 + a4)]t [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] [(B1 + B3 + Bs) + A (B2 + B4)] [((B2 + B4) – (B1 + B3 + B5)) (A+1)] | = 0, (4.19)
Since P+ Q, it follows that [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) + A (82 + 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 all P,Q E R, we have [(a1 + a3 + a5) - (a2+ a4)] [(B1 + B3 + B3) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] [(B1 + B3 + B5) + A (B2 + B4)] [((B2 + Ba) – (B1 + B3 + B3)) (A+1)] PQ (4.18) Let P and Q are two distinct real roots of the quadratic equation t - (P+Q)t+ PQ = 0. %3D [(B1 + B3 + B5) + A (B2 + B4)] t² – [(a1 + a3 + a5) –- (a2 + a4)]t [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] [(B1 + B3 + Bs) + A (B2 + B4)] [((B2 + B4) – (B1 + B3 + B5)) (A+1)] | = 0, (4.19)
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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Topic Video
Question
Explain the determine yellow and the inf is here
![Theorem 6 If (a1 +az + 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 + B5) (a2 +a4) + A (B2 + B4) (a1 + az + 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 +¤4Ym–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 + B3) 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 + Bз + Bs) Р? + (B2 + Ba) PQ
A (B1 + B3 + B5) PQ + A (B2 + B4) Q²
+ (a1 + a3 + a5) P+(a2+a4) Q,
(4.15)
and
(B1 + B3 + Bs) Q² + (B2 + B4) PQ = A(B1 + 3 + 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 + 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)I
PQ
[(B1 + B3 + B5) + 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
t? - (P+ Q)t + PQ = 0.
[(B1 + B3 + B5) + A (B2 + B4)] t² – [(a1 + az + a5) – (a2 + a4)]t
[(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)]
|
[(B1 + B3 + B5) + A (B2 + B4)] [((B2 + B4) – (B1 + B3 + Bs)) (A+ 1)]
0,
(4.19)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe1c5a833-e8e1-4192-bf76-b19122b6605c%2F63285442-dfd0-4406-9049-616306221e18%2Farj9y9m_processed.png&w=3840&q=75)
Transcribed Image Text:Theorem 6 If (a1 +az + 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 + B5) (a2 +a4) + A (B2 + B4) (a1 + az + 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 +¤4Ym–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 + B3) 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 + Bз + Bs) Р? + (B2 + Ba) PQ
A (B1 + B3 + B5) PQ + A (B2 + B4) Q²
+ (a1 + a3 + a5) P+(a2+a4) Q,
(4.15)
and
(B1 + B3 + Bs) Q² + (B2 + B4) PQ = A(B1 + 3 + 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 + 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)I
PQ
[(B1 + B3 + B5) + 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
t? - (P+ Q)t + PQ = 0.
[(B1 + B3 + B5) + A (B2 + B4)] t² – [(a1 + az + a5) – (a2 + a4)]t
[(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)]
|
[(B1 + B3 + B5) + A (B2 + B4)] [((B2 + B4) – (B1 + B3 + Bs)) (A+ 1)]
0,
(4.19)
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
Ym+1 =
Aym+
т 3D 0, 1, 2, ...,
В1ут-1 + В2ут-2 + Взут-3 + Влут-4 + B5ут-5
(1.1)
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