[(a + 1) ((d + g) – (e+s))] [(b+ f) – (c+r)]² +4 [(b+ f) – (c+r)] [(c+r) (d + g) + a (e + s) (b + f)] > 0. - |
[(a + 1) ((d + g) – (e+s))] [(b+ f) – (c+r)]² +4 [(b+ f) – (c+r)] [(c+r) (d + g) + a (e + s) (b + f)] > 0. - |
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 got eq13 from eq1 explain this step please
Transcribed Image Text:The objective of this article is to investigate some qualitative behavior of
the solutions of the nonlinear difference equation
bxn-1 + cxn-2 + fxn-3 +rxn-4
Xn+1 = axn +
n = 0, 1, 2, .. (1)
dxn-1+ exn-2+ gxn-3 + sxn-4
where the coefficients a b cde fa rs E (0 o) while the initial con-
![Theorem 6 If (b+f) > (c+r) and (d + g) > (e + s), then the necessary
and sufficient condition for Eq. (1) to have positive solutions of prime period
two is that the inequality
[(a + 1) ((d + g) – (e+ s))] [(b+ f) – (c+r)]?
+4 [(b+ f) – (c+ r)] [(c+r) (d + g) + a (e + s) (b + f)] > 0. (13)
is valid.
Proof: Suppose that there exist positive distinctive solutions of prime period
two
P,Q, P, Q, ..
of Eq.(1). From Eq.(1) we have
bxn-1+ cxn-2+ fxn-3 +rxn-4
Xn+1 = axn +
dxn-1 + exп-2 + grn-з + stn-4
(ь + f) Р+ (с+ r) Q
(d +g) P + (e + s) Q'
(b+ f) Q+ (c+r) P
P = aQ+
Q = aP+
(d + g) Q+ (e+ s) P'
Consequently, we obtain
(d + g) P² + (e + s) PQ = a (d + g) PQ+a(e+s) Q² + (b+ f) P+(c+r) Q,
(14)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F25e1145d-beeb-4bcc-a46c-2420650cb54d%2Fa46140de-14be-4083-8f3b-1dd6ea557166%2Fcu9j7rk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 6 If (b+f) > (c+r) and (d + g) > (e + s), then the necessary
and sufficient condition for Eq. (1) to have positive solutions of prime period
two is that the inequality
[(a + 1) ((d + g) – (e+ s))] [(b+ f) – (c+r)]?
+4 [(b+ f) – (c+ r)] [(c+r) (d + g) + a (e + s) (b + f)] > 0. (13)
is valid.
Proof: Suppose that there exist positive distinctive solutions of prime period
two
P,Q, P, Q, ..
of Eq.(1). From Eq.(1) we have
bxn-1+ cxn-2+ fxn-3 +rxn-4
Xn+1 = axn +
dxn-1 + exп-2 + grn-з + stn-4
(ь + f) Р+ (с+ r) Q
(d +g) P + (e + s) Q'
(b+ f) Q+ (c+r) P
P = aQ+
Q = aP+
(d + g) Q+ (e+ s) P'
Consequently, we obtain
(d + g) P² + (e + s) PQ = a (d + g) PQ+a(e+s) Q² + (b+ f) P+(c+r) Q,
(14)
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