[(a1 + az + a5) – (a2 + a4)] + & 2 [(B1 + B3 + B5) + A (B2 + B4)] (4.21) [(a1 + a3 + a5) - (a2+ a4)] – 6 2 [(B1 + B3 + B5) + A (B2 + B4)] (4.22) 8 = V[(a1 + a3 + a5) – (a2 + a4)]² – n,
[(a1 + az + a5) – (a2 + a4)] + & 2 [(B1 + B3 + B5) + A (B2 + B4)] (4.21) [(a1 + a3 + a5) - (a2+ a4)] – 6 2 [(B1 + B3 + B5) + A (B2 + B4)] (4.22) 8 = V[(a1 + a3 + a5) – (a2 + a4)]² – n,
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
Show me the steps of determine blue and inf is here please step by step and i want every details
![[(a1 + a3 + a5) - (a2 + a4)] + ô
P =
(4.21)
2 [(81 + B3 + B5) + A (B2 + B4)]
and
[(a1 + az + a5) – (a2 + a4)] – 6
Q =
(4.22)
2 [(B1 + B3 + B5) + A (B2 + B4)]
where
8 = V[(a1 + a3 + a5) – (a2 + a4)]² – n,
and
4[(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (aj + az + 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
a1y-1+ a2y-2+ a3y-3 +a4Y-4 + a5Y-5
Y1 =
Ayo +
Biy-1 + B2y-2 + B3y-3 + B4y-4 + Bsy-5
(a1 + a3 + a5) P + (a2 + a4) Q
= AQ +
(B1 + B3 + B3) P + (B2 + B4) Q
(4.23)
||
The main focus of this article is to discuss some qualitative behavior of
the solutions of the nonlinear difference equation
a1Ym-1 + a2Ym-2 + a3Ym-3 + a4Ym-4 + a5Ym-5
Aym+
Вут-1 + В2ут-2 + Взут-3 + В43/m -4 + Взут-5
т 3D 0, 1, 2, ..,
Ym+1 =
(1.1)
where the coefficients A, a;, Bi E (0, 00), i = 1, ..., 5, while the initial condi-
Y-5,Y-4,y-3,Y-2,Y-1, Yo are arbitrary positive real numbers. Note that
the special case of Eq.(1.1) has been discussed in [4] when az = B3 = a4 =
= a5 = B5 = 0 and Eq.(1.1) has been studied in [8] in the special case
tions
B4
when a4 =
B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the
special case when as = B5 = 0.
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F5f17b113-66c2-4cf7-b464-826fa4ed3ce3%2F606zegb_processed.png&w=3840&q=75)
Transcribed Image Text:[(a1 + a3 + a5) - (a2 + a4)] + ô
P =
(4.21)
2 [(81 + B3 + B5) + A (B2 + B4)]
and
[(a1 + az + a5) – (a2 + a4)] – 6
Q =
(4.22)
2 [(B1 + B3 + B5) + A (B2 + B4)]
where
8 = V[(a1 + a3 + a5) – (a2 + a4)]² – n,
and
4[(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (aj + az + 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
a1y-1+ a2y-2+ a3y-3 +a4Y-4 + a5Y-5
Y1 =
Ayo +
Biy-1 + B2y-2 + B3y-3 + B4y-4 + Bsy-5
(a1 + a3 + a5) P + (a2 + a4) Q
= AQ +
(B1 + B3 + B3) P + (B2 + B4) Q
(4.23)
||
The main focus of this article is to discuss some qualitative behavior of
the solutions of the nonlinear difference equation
a1Ym-1 + a2Ym-2 + a3Ym-3 + a4Ym-4 + a5Ym-5
Aym+
Вут-1 + В2ут-2 + Взут-3 + В43/m -4 + Взут-5
т 3D 0, 1, 2, ..,
Ym+1 =
(1.1)
where the coefficients A, a;, Bi E (0, 00), i = 1, ..., 5, while the initial condi-
Y-5,Y-4,y-3,Y-2,Y-1, Yo are arbitrary positive real numbers. Note that
the special case of Eq.(1.1) has been discussed in [4] when az = B3 = a4 =
= a5 = B5 = 0 and Eq.(1.1) has been studied in [8] in the special case
tions
B4
when a4 =
B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the
special case when as = B5 = 0.
%3D
![Substituting (4.21) and (4.22) into (4.23) we deduce that
(@1 + a3 + a5) P+ (a2+ a4) Q
(B1 + B3 + B5) P + (B2 + B4) Q
[A (B1 + B3 + B5) – (B2 + B4)]PQ + A (B2 + B4) Q² – (B1 + B3 + B5) P²
Yı – P
= AQ +
(B1 + B3 + B5) P+ (52 + B4) Q
(a1 + a3 + a5) P+(a2 + a4) Q
(B1 + B3 + B5) P+(B2 + B4) Q
+
[A(31+B3+B5)-(B2+B4)][S1][(B1+B3+B5)(a2+a4)+A(B2+B4) (a1+a3+a5)]
[(B1+B3+B5)+A(ß2+B4)]²[((B2+B4)–(B1+B3+B5))(A+1)]
(B1 + B3 + B5)
[(@1+a3+a5)-(a2+a4)]+8
2[(B1+B3+B5)+A(B2+B4)]
+ (B2 + B4)
[(a1+a3+a5)-(a2+a4)]-8
2[(B1+B3+B5)+A(B2+B4)]
A (B2 + B4) (-
2
[(a1+a3+a5)-(a2+a4)]-8
2[(B1+B3+B5)+A(B2+B4)]
(ß1 + B3 + B5)
2
[(@1+a3+a5)-(a2+a4)]+8
2[(B1+B3+B5)+A(B2+B4)]
+
(B1 + B3 + B5)
[(a1+a3+a5)-(a2+a4)]+8
2[(B1+33+B5)+A(B2+B4)]
+ (B2 + B4) ( ai+a3+a5)-(a2+a4)]-8
2[(B1+B3+B5)+A(B2+B4)]
[(@1+a3+a5)-(a2+a4)]-ô
2[(B1+B3+B5)+A(B2+B4)]
[(a1+a3+a5)-(@2+a4)]-8
2[(B1+B3+B5)+A(B2+B4)]
(a1 + a3 + a5) ( aita3+a5)-(a2+a4)]+8
+ (a2 + a4)
2[(B1+B3+B5)+A(B2+B4)]
[(@1+a3+a5)-(a2+a4)]+8
2[(B1+B3+B5)+A(ß2+B4)]
+
(B1 + B3 + B5)
+ (B2 + B4)
(4.24)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F5f17b113-66c2-4cf7-b464-826fa4ed3ce3%2Fl564at_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Substituting (4.21) and (4.22) into (4.23) we deduce that
(@1 + a3 + a5) P+ (a2+ a4) Q
(B1 + B3 + B5) P + (B2 + B4) Q
[A (B1 + B3 + B5) – (B2 + B4)]PQ + A (B2 + B4) Q² – (B1 + B3 + B5) P²
Yı – P
= AQ +
(B1 + B3 + B5) P+ (52 + B4) Q
(a1 + a3 + a5) P+(a2 + a4) Q
(B1 + B3 + B5) P+(B2 + B4) Q
+
[A(31+B3+B5)-(B2+B4)][S1][(B1+B3+B5)(a2+a4)+A(B2+B4) (a1+a3+a5)]
[(B1+B3+B5)+A(ß2+B4)]²[((B2+B4)–(B1+B3+B5))(A+1)]
(B1 + B3 + B5)
[(@1+a3+a5)-(a2+a4)]+8
2[(B1+B3+B5)+A(B2+B4)]
+ (B2 + B4)
[(a1+a3+a5)-(a2+a4)]-8
2[(B1+B3+B5)+A(B2+B4)]
A (B2 + B4) (-
2
[(a1+a3+a5)-(a2+a4)]-8
2[(B1+B3+B5)+A(B2+B4)]
(ß1 + B3 + B5)
2
[(@1+a3+a5)-(a2+a4)]+8
2[(B1+B3+B5)+A(B2+B4)]
+
(B1 + B3 + B5)
[(a1+a3+a5)-(a2+a4)]+8
2[(B1+33+B5)+A(B2+B4)]
+ (B2 + B4) ( ai+a3+a5)-(a2+a4)]-8
2[(B1+B3+B5)+A(B2+B4)]
[(@1+a3+a5)-(a2+a4)]-ô
2[(B1+B3+B5)+A(B2+B4)]
[(a1+a3+a5)-(@2+a4)]-8
2[(B1+B3+B5)+A(B2+B4)]
(a1 + a3 + a5) ( aita3+a5)-(a2+a4)]+8
+ (a2 + a4)
2[(B1+B3+B5)+A(B2+B4)]
[(@1+a3+a5)-(a2+a4)]+8
2[(B1+B3+B5)+A(ß2+B4)]
+
(B1 + B3 + B5)
+ (B2 + B4)
(4.24)
Expert Solution
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