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)]² +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+ fan-3+ rXn-4 Xn+1 = axn + %3D dxn-1 + exn-2+ gan-3 + sxn-4 (b+ f) P+ (c+r)Q (d + g) P + (e + s)Q' (b+ f)Q + (c+ r) P (d + g) Q + (e + s) P' P = aQ+ Q = aP+ Consequently, we obtain (d+ g) P² + (e + s) PQ = a (d + g) PQ+a(e+ s) Q² + (b + f)P+(c+r)Q, (14) %3D
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)]² +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+ fan-3+ rXn-4 Xn+1 = axn + %3D dxn-1 + exn-2+ gan-3 + sxn-4 (b+ f) P+ (c+r)Q (d + g) P + (e + s)Q' (b+ f)Q + (c+ r) P (d + g) Q + (e + s) P' P = aQ+ Q = aP+ Consequently, we obtain (d+ g) P² + (e + s) PQ = a (d + g) PQ+a(e+ s) Q² + (b + f)P+(c+r)Q, (14) %3D
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Explain the determine grren
![l stc ksa
9:21 PM
C@ 974%4
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 = aan +
dxn-1 + exn-2+ gxn-3 + sxn-4
(P = aQ+ d+9) P + (e + s) Q'
(b+ f)Q + (c+ r) P
(d + g) Q + (e + s) P'
Q = aP+
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%2F86116bc7-59c5-4d89-8ef1-07148d796d94%2Ff0f9040f-eb8f-42d1-a828-91cfeb476181%2F43xic0i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:l stc ksa
9:21 PM
C@ 974%4
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 = aan +
dxn-1 + exn-2+ gxn-3 + sxn-4
(P = aQ+ d+9) P + (e + s) Q'
(b+ f)Q + (c+ r) P
(d + g) Q + (e + s) P'
Q = aP+
Consequently, we obtain
(d+g) P² + (e + s) PQ = a (d + g) PQ+a (e+ s) Q² + (b+ f) P+(c+r) Q,
(14)
Transcribed Image Text:l stc ksa
12:14 AM
@ 1 60% 4
The objective of this article is to investigate some qualitative behavior of
the solutions of the nonlinear difference equation
bxn-1 + cæn-2+ fxn-3 + ræn-4
Xn+1 = axn +
n = 0, 1, 2, . (1)
dxn-1+ exn-2 + gæn-3 + sxn-4
where the coefficients a, b, c, d, e, f, g, r, s E (0, 00), while the initial con-
ditions a_4,x_3,x_2, x-1, xo are arbitrary positive real numbers. Note that
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