(d + g) Q² + (e + s) PQ = a (d + g) PQ+a(e+s) P² + (b+ f) Q+(c+ r) P. (15) By subtracting (14) from (15), we have [a (e + s) + (d + g)] (P² – Q²) = [(b+ f) – (c+r)] (P – Q). %3D Since P + Q, it follows that [(b + f) – (c+r)] [a (e + s) + (d + g)]' P+Q = (16)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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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 + fxn-3 + rxn-4
dxn-1 + exn-2+ gxn-3 + sxn-4
Xn+1 = axn +
(b + f) P+ (c+r) Q
(d+ g) P+ (e + s) Q'
(6+ 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)
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)]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 + fxn-3 + rxn-4 dxn-1 + exn-2+ gxn-3 + sxn-4 Xn+1 = axn + (b + f) P+ (c+r) Q (d+ g) P+ (e + s) Q' (6+ 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)
and
(d + g) Q² + (e + s) PQ = a (d + g) PQ + a (e + s) P² +(b+ f) Q+(c+ r) P.
(15)
By subtracting (14) from (15), we have
[a (e + s) + (d + g)) (P² – Q²) = [(b+ f) – (c+r)] (P – Q).
Since P + Q, it follows that
[(b+ f) – (c+r)]
[a (e+ s) + (d+ g)]'
P+Q =
(16)
while, by adding (14) and (15) and by using the relation
p² + Q? = (P+Q)² – 2PQ for all
P, Q E R,
we obtain
Transcribed Image Text:and (d + g) Q² + (e + s) PQ = a (d + g) PQ + a (e + s) P² +(b+ f) Q+(c+ r) P. (15) By subtracting (14) from (15), we have [a (e + s) + (d + g)) (P² – Q²) = [(b+ f) – (c+r)] (P – Q). Since P + Q, it follows that [(b+ f) – (c+r)] [a (e+ s) + (d+ g)]' P+Q = (16) while, by adding (14) and (15) and by using the relation p² + Q? = (P+Q)² – 2PQ for all P, Q E R, we obtain
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