Theorem 6 If (b+ f) > (c+r) und (d+g) > (e+ s), then the necessary und 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 bæn-1+ can-2+ fæn-3 + rxn-4 In+1 = axn+ dæn-1 + exn-2+ gæn-3 + sæn-4 (6+ 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)Q?+(b+ f) P+(c+r)Q, (14) and (d + g) Q? + (e+ s) PQ = a (d+ g) PQ+a(e+s) P2 + (b6+ f)Q+(c+r) P. (15) By subtracting (14) from (15), we have (a (e + s) + (d+ g)I (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) while, by adding (14) and (15) and by using the relation p² + Q? = (P+ Q)² – 2PQ for all P,Q ER, %3D we obtain [(b+ f) – (c+r)] [(c+r) (d+ g) + a (e +s) (b+ f)] (17) PQ = [a (e + s) + (d+ g)] [((e+ s) - (d+ g)) (a + 1)] Assume that P and Q are two distinct real roots of the quadratic equation t2 - (P+Q)t+ PQ = 0. (a (e + s) + (d+ g)] t² – [(b+ f) – (c+r)] t [(b+ f) - (c+r)] [(c+r) (d+ g) + a (e + s) (b+f)] [a (e+ s) + (d+ g)] [(e+ s) - (d+ g)) (a + 1)] 0, (18) and so - (c+r))? – 4 (b +f)- (c+r)][(c+r) (d + g) + a (e + s) (b+f)] [(a + 1) ((e + s) - (d+g))] >0, or + S) – (c+r))² + 4 [(b+ f) – (c+r)[(c+r) (d + g) + a (e + s) (b+ f)] [(a +1) ((d + g) –- (e+s))] >0. (19) From (19), we obtain [(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. Thus, the condition (13) is valid. Conversely, suppose that the condition (13) is valid where (b+ f) > (c+r) and (d+g) > (e+ s). Then, we deduce
Theorem 6 If (b+ f) > (c+r) und (d+g) > (e+ s), then the necessary und 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 bæn-1+ can-2+ fæn-3 + rxn-4 In+1 = axn+ dæn-1 + exn-2+ gæn-3 + sæn-4 (6+ 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)Q?+(b+ f) P+(c+r)Q, (14) and (d + g) Q? + (e+ s) PQ = a (d+ g) PQ+a(e+s) P2 + (b6+ f)Q+(c+r) P. (15) By subtracting (14) from (15), we have (a (e + s) + (d+ g)I (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) while, by adding (14) and (15) and by using the relation p² + Q? = (P+ Q)² – 2PQ for all P,Q ER, %3D we obtain [(b+ f) – (c+r)] [(c+r) (d+ g) + a (e +s) (b+ f)] (17) PQ = [a (e + s) + (d+ g)] [((e+ s) - (d+ g)) (a + 1)] Assume that P and Q are two distinct real roots of the quadratic equation t2 - (P+Q)t+ PQ = 0. (a (e + s) + (d+ g)] t² – [(b+ f) – (c+r)] t [(b+ f) - (c+r)] [(c+r) (d+ g) + a (e + s) (b+f)] [a (e+ s) + (d+ g)] [(e+ s) - (d+ g)) (a + 1)] 0, (18) and so - (c+r))? – 4 (b +f)- (c+r)][(c+r) (d + g) + a (e + s) (b+f)] [(a + 1) ((e + s) - (d+g))] >0, or + S) – (c+r))² + 4 [(b+ f) – (c+r)[(c+r) (d + g) + a (e + s) (b+ f)] [(a +1) ((d + g) –- (e+s))] >0. (19) From (19), we obtain [(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. Thus, the condition (13) is valid. Conversely, suppose that the condition (13) is valid where (b+ f) > (c+r) and (d+g) > (e+ s). Then, we deduce
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
![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+ can-2 + fæn-3 + ræn-4
En+1 = axn +
dæn-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)
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
[(b+ f) – (c+r)I [(c+r) (d+ g) + a (e+ s) (b+ f)]
PQ =
(17)
(a (e + s) + (d + g)]² [((e+ s) – (d+ g)) (a + 1)]
Assume that P and Q are two distinct real roots of the quadratic equation
t2 - (P+Q)t + PQ = 0.
[a (e + s) + (d + g)] t? – [(b+ f) – (c+r)] t
[(b+ f) – (c+r)] [(c+r) (d+ g) + a (e + s) (b+ f)]
[a (e + s) + (d + g9)) [((e + s) – (d + g)) (a + 1)]
0,
(18)
and so
((6+) – (c+r))? _ 4[(b+ f) – (c + r)][(c+r) (d+g) + a (e +s) (b+f)]
> 0,
[(a + 1) ((e+ s) – (d+ g))]
or
(6+ ) – (c+r))? + 4[(b+ f) – (c+ r)] [(e +r) (d + g) + a (e + s) (b + f)]
> 0.
[(a + 1) ((d+ g) – (e+ s))]
(19)
From (19), we obtain
[(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.
Thus, the condition (13) is valid. Conversely, suppose that the condition
(13) is valid where (b+ f) > (c+r) and (d+g) > (e+ s). Then, we deduce](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb06891c1-7310-49aa-a044-9ddf722456d6%2F0678c82c-971a-4ef2-beed-f40b1bc62d88%2Fzxkq6k0b_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+ can-2 + fæn-3 + ræn-4
En+1 = axn +
dæn-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)
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
[(b+ f) – (c+r)I [(c+r) (d+ g) + a (e+ s) (b+ f)]
PQ =
(17)
(a (e + s) + (d + g)]² [((e+ s) – (d+ g)) (a + 1)]
Assume that P and Q are two distinct real roots of the quadratic equation
t2 - (P+Q)t + PQ = 0.
[a (e + s) + (d + g)] t? – [(b+ f) – (c+r)] t
[(b+ f) – (c+r)] [(c+r) (d+ g) + a (e + s) (b+ f)]
[a (e + s) + (d + g9)) [((e + s) – (d + g)) (a + 1)]
0,
(18)
and so
((6+) – (c+r))? _ 4[(b+ f) – (c + r)][(c+r) (d+g) + a (e +s) (b+f)]
> 0,
[(a + 1) ((e+ s) – (d+ g))]
or
(6+ ) – (c+r))? + 4[(b+ f) – (c+ r)] [(e +r) (d + g) + a (e + s) (b + f)]
> 0.
[(a + 1) ((d+ g) – (e+ s))]
(19)
From (19), we obtain
[(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.
Thus, the condition (13) is valid. Conversely, suppose that the condition
(13) is valid where (b+ f) > (c+r) and (d+g) > (e+ s). Then, we deduce
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