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[(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
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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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
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)
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
Step 1

Given,        P=α1+α3+α5-α2+α4+δ2β1+β3+β5+Aβ2+β4                ...(4.21)        Q=α1+α3+α5-α2+α4-δ2β1+β3+β5+Aβ2+β4                ...(4.22)where            δ=α1+α3+α5-α2+α42-ηand          η=4α1+α3+α5-α2+α4β1+β3+β5α2+α4+Aβ2+β4α1+α3+α5β2+β4-β1+β3+β5(A+1)and       y1=AQ+α1+α3+α5P+α2+α4Qβ1+β3+β5P+β2+β4Q           ...(4.23)

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