(ai + a3 + a5) P+ (@2+ a4) Q (B1 + B3 + B3) P+ (B2 + B4) Q [A (B1 + B3 + B5)- (B2+ B4)]PQ + A (82 + B4) Q2 - (B1 + B3 + B5) P2 Y1 – P = AQ + - P | | (B1 + B3 + B5) P + (B2 + B4) Q (a1 + a3 + a5)P+ (a2+ a4) Q + (B1 + B3 + B5) P+ (82+ B4) Q
(ai + a3 + a5) P+ (@2+ a4) Q (B1 + B3 + B3) P+ (B2 + B4) Q [A (B1 + B3 + B5)- (B2+ B4)]PQ + A (82 + B4) Q2 - (B1 + B3 + B5) P2 Y1 – P = AQ + - P | | (B1 + B3 + B5) P + (B2 + B4) Q (a1 + a3 + a5)P+ (a2+ a4) Q + (B1 + B3 + B5) P+ (82+ B4) Q
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 red and the inf is here
![There exist two positive distinctive real numbers P and Q representing two
positive roots of Eq.(4.19) such that
[(a1 + a3 + a5) – (a2 + a4)] + 8
P =
(4.21)
2 [(B1 + B3 + B5) + A (B2 + B4)]
and
[(a1 + a3 + a5) – (a2 + a4)] – 8
2 [(81 + B3 + B5) + A (82 + B4)]
(4.22)
where
8 = V[(a1 + a3 +a5) – (a2 + a4)]² – n,
and
4 [(a1+ a3 + a5) – (a2 + a4)] [(ß1 + B3 + B5) (œ2 + a4) + A (B2 + B4) (a1 + a3 + a5)]
[((32 + B4) – (B1 + B3 + B5)) (A + 1)]
Now, let us prove that P and Q are positive solutions of prime period two of
Eq.(1.1). To this end, we assume that y-5 = P, y-4 = Q, y–3 = P, y-2 =
Q, y-1 = P, Yo= Q. Now, we are going to show that Yı = P and y2 = Q.
From Eq.(1.1) we deduce that
а1у-1 + ӕ2у-2 + азу-з + аду-4 + 05у-5
Y1 = Ayo +
B1y-1 + B2y-2 + B3y-3 + B4Y-4 + B5Y-5
(@1 + a3 + a5) P+ (a2+ a4) Q
= AQ +
(4.23)
(B1 + Вз + В5) Р + (82 + Вa) Q](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe1c5a833-e8e1-4192-bf76-b19122b6605c%2Fe9167276-67c6-4f29-a000-db13d6277364%2Fd09jdu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:There exist two positive distinctive real numbers P and Q representing two
positive roots of Eq.(4.19) such that
[(a1 + a3 + a5) – (a2 + a4)] + 8
P =
(4.21)
2 [(B1 + B3 + B5) + A (B2 + B4)]
and
[(a1 + a3 + a5) – (a2 + a4)] – 8
2 [(81 + B3 + B5) + A (82 + B4)]
(4.22)
where
8 = V[(a1 + a3 +a5) – (a2 + a4)]² – n,
and
4 [(a1+ a3 + a5) – (a2 + a4)] [(ß1 + B3 + B5) (œ2 + a4) + A (B2 + B4) (a1 + a3 + a5)]
[((32 + B4) – (B1 + B3 + B5)) (A + 1)]
Now, let us prove that P and Q are positive solutions of prime period two of
Eq.(1.1). To this end, we assume that y-5 = P, y-4 = Q, y–3 = P, y-2 =
Q, y-1 = P, Yo= Q. Now, we are going to show that Yı = P and y2 = Q.
From Eq.(1.1) we deduce that
а1у-1 + ӕ2у-2 + азу-з + аду-4 + 05у-5
Y1 = Ayo +
B1y-1 + B2y-2 + B3y-3 + B4Y-4 + B5Y-5
(@1 + a3 + a5) P+ (a2+ a4) Q
= AQ +
(4.23)
(B1 + Вз + В5) Р + (82 + Вa) Q
![Substituting (4.21) and (4.22) into (4.23) we deduce that
АQ +
(с1 + аз + as)Р+ (а2 + а4) Q
- P
У1 — Р —
(B1 + Вз + B5) Р+ (32 + Ва) Q
[A (B1 + B3 + B3) – (B2 + Ba)]PQ + A (B2 + B4) Q² – (B1 + ß3 + B5) P²
(B1 + B3 + B5) P+(B2+ B4) Q
(a1 + a3 + a5) P+(@2+ a4) Q
(B1 + B3 + B5) P+(82+ B4) Q
([A(B1+33+Bs)-(82+B4)][S1][(B1+33+B5)(@2+a4)+A(82+B4) (ai+a3+a5)]
[(81+33+B5)+A(82+34)]²[((82+B4)-(ß1+B3+B5))(A+1)]
(B1 + B3 + B5)
[(a1+a3+a5)-(a2+a4)]+ô
2[(B1+B3+B5)+A(B2+B4)]
+ (B2 + B1) (aita3tas)-(a2ta4)]-8
2[(B1+B3+B3)+A(82+B4)]
2
[(@1+a3+a5)-(a2+a4)]-8
A (B2 + B4) ( 2(31+83+Bs)+A(B2+B4)]
(a1ta3+as)-(a2ta4)]+ô\²
2[(31+33+ßs)+A(B2+B4)]
(B1 + B3 + B5)
+ (B2 + Ba) ( laita3+a5)-(ag+a4)]-8)
2[(B1+83+B5)+A(B2+B4)]
[(@1+a3+ag)-(a2+a4)]+8`
2[(81+B3+B5)+A(B2+B4)] .
[(a1+a3+a5)-(a2+a4)]+8`
2[(B1+33+B5)+A(32+B4)]
(B1 + B3 + B5)
(a1 + a3 + a5)
+
+ (a2 + a4)
[(a1+a3+a5)-(a2+a4)]-ô
2[(81+B3+B3)+A(B2+B4)] )
(B1 + B3 + B5)
[(a1ta3+a5)-(a2+a4)]+8
2[(B1+33+B5)+A(82+B4)]
+ (B2 + Ba) ( ata3+a;)-(ag+a4)]-8
2[(31+B3+Bs)+A(B2+B4)]
(4.24)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe1c5a833-e8e1-4192-bf76-b19122b6605c%2Fe9167276-67c6-4f29-a000-db13d6277364%2Fyjplxo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Substituting (4.21) and (4.22) into (4.23) we deduce that
АQ +
(с1 + аз + as)Р+ (а2 + а4) Q
- P
У1 — Р —
(B1 + Вз + B5) Р+ (32 + Ва) Q
[A (B1 + B3 + B3) – (B2 + Ba)]PQ + A (B2 + B4) Q² – (B1 + ß3 + B5) P²
(B1 + B3 + B5) P+(B2+ B4) Q
(a1 + a3 + a5) P+(@2+ a4) Q
(B1 + B3 + B5) P+(82+ B4) Q
([A(B1+33+Bs)-(82+B4)][S1][(B1+33+B5)(@2+a4)+A(82+B4) (ai+a3+a5)]
[(81+33+B5)+A(82+34)]²[((82+B4)-(ß1+B3+B5))(A+1)]
(B1 + B3 + B5)
[(a1+a3+a5)-(a2+a4)]+ô
2[(B1+B3+B5)+A(B2+B4)]
+ (B2 + B1) (aita3tas)-(a2ta4)]-8
2[(B1+B3+B3)+A(82+B4)]
2
[(@1+a3+a5)-(a2+a4)]-8
A (B2 + B4) ( 2(31+83+Bs)+A(B2+B4)]
(a1ta3+as)-(a2ta4)]+ô\²
2[(31+33+ßs)+A(B2+B4)]
(B1 + B3 + B5)
+ (B2 + Ba) ( laita3+a5)-(ag+a4)]-8)
2[(B1+83+B5)+A(B2+B4)]
[(@1+a3+ag)-(a2+a4)]+8`
2[(81+B3+B5)+A(B2+B4)] .
[(a1+a3+a5)-(a2+a4)]+8`
2[(B1+33+B5)+A(32+B4)]
(B1 + B3 + B5)
(a1 + a3 + a5)
+
+ (a2 + a4)
[(a1+a3+a5)-(a2+a4)]-ô
2[(81+B3+B3)+A(B2+B4)] )
(B1 + B3 + B5)
[(a1ta3+a5)-(a2+a4)]+8
2[(B1+33+B5)+A(82+B4)]
+ (B2 + Ba) ( ata3+a;)-(ag+a4)]-8
2[(31+B3+Bs)+A(B2+B4)]
(4.24)
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