Show me the steps of determine green and inf is here (a1 + a3 + a5) P+(a2+a4) QУл — Р 3D AQ +- P(B1 + B3 + B5) P + (B2 + B4) Q[A (B1 + B3 + B6) – (B2 + B4)]PQ + A (B2 + B4) Q² – (B1 + Ba3 + Bs) P²(B1 + B3 + B3) P + (B2 + B4) Q(a1 + a3 + a5) P + (@2+a4)Q(В + Вз + В5) Р + (В2 + Ba) Q[A(B1+B3+Bs)-(82+B4)][S1][(B1+B3+Bs)(a2+a4)+A(B2+B4) (a1+a3+a5)][(B1+B3+Bs)+A(B2+B4)]*[((B2+B4)-(B1+B3+B5))(A+1)]([(a1+a3+a5)-(a2+a4)]+8(B1 + B3 + B5) (2(31+B3+Bs)+A(B2+Ba)]+ (B2 + B4) (Citag+a5)-(a9+as)]-62((B1+B3+B5)+A(B2+B4)]22A (B2 + B4) ((@i+a3+as)-(az+as)]-62[(B1+B3+B5)+A(B2+B4)]-(81 + B3 + Bs)[(@1+a3+as)-(a2+as)]+82(B1+B3+Bs)+A(ß2+B4)]PI T P3 + P5) (231+33+85)+A(62+8) + (B2 + B4) (eIta3+as)-(a2+a4)]-8[(ai+a3+as)-(a2+a4)]-82[(B1+B3+B5)+A(B82+B4)]I(a1+a3+as)-(a2+a4)]-82((B1+33+B5)+A(B2+B4)(4.24)[(ai+a3+a5)-(a2+a4)]+82[(B1+B3+Bs)+A(B2+B4)](a1+a3+as)-(a2+a4)]+ô2[(B1+B3+B5)+A(B2+B4)](a1+a3+as)-(aztas)]+82[(B1+B3+B3)+A(B2+B4)](a1 + a3 + a5)+ (a2 + a4)(B1 + B3 + B5)+ (B2 + B4)The main focus of this article is to discuss some qualitative behavior ofthe solutions of the nonlinear difference equationa1Ym-1+a2Ym-2 + a3Ym-3 + a4Ym-4 + a5Ym-5Ym+1 =Aym+m = 0,1,2, ..,В1ут-1 + В2ут-2 + Взут-3 + Ваут-4 + Bзут-5(1.1)where the coefficients A, a;, Bi E (0, 00), i = 1, .., 5, while the initial condi-tions y-5,y-4,y-3,y-2,y-1, yo are arbitrary positive real numbers. Note thatthe special case of Eq.(1.1) has been discussed in [4] when az =B4when a4 == a4 == 0 and Eq.(1.1) has been studied in [8] in the special caseB4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the= a5 == B5special case when as = B5 = 0. Multiplying the denominator and numerator of (4.24) by 4 [(B1 + B3 + B5) + A (B2 + B4)]?we get4[A(B1+B3+B5)-(B2+B4)][S1][(B1+ß3+B5)(@2+a4)+A(B2+B4) (a1+a3+æ5)][(B2+B4)-(B1+B3+Bs))(A+1)]Y1 – P =A (B2 + B4) (S1 – 8)² – (B1 + B3 + Bs) (Sı + 8)²S2[(B1 + B3 + B5) + A (B2 + B4)] (aı + a3 + a5) (S1 + 8)+S2[(B1 + B3 + B5) + A (B2 + B4)] (a2 + a4) (S1 – 8)+S4[A(B1+B3+B5)-(B2+Ba)][S1][(B1+B3+Bs)(a2+a4)+A(B2+B4) (a1+a3+a5)][((32+B4)-(B1+ß3+B5))(A+1)]A (B2 + Ba) [S1]? - (B1 + B3 + B5) [S1]?[A (B2 + B4) – (ß1 + B3 + B3)]82S142[(B1 + B3 + B5) + A (B2 + B4)] (a2+a4) [S1]S2[(B1 + B3 + B5)+ A(B2 + B4)] (a1 + a3 + a5) [S1]S2A (B2 + B4) [(a1+az + a5) – (a2 + a4)] & + 2 (B1 + B3 + B5) [S1] &S2[(B1 + B3 + B5) + A (B2 + Ba)] (a1+ a3 + a5) 8 – 2[(B1 + B3 + B5) + A (B2 + Ba)] (a2 + a4) 8S4[A(B1+B3+B5)-(B2+B4)][S1][(B1+B3+B5)(a2+a4)+A(B2+B4)(a1+a3+as)][((B2+B4)-(B1+B3+B5))(A+1)]S4[Si][(B1+B3+ßs)(a2+as)+A(ß2+ß4)(@1+a3+as)][A(B2+Ba)-(B1+B3+ßs)1[((B2+B4)-(B1+ß3+B5))(A+1)]A (B2 + B4) [S1]? – (B1 + B3 + B3) [S1]²-S[A (B2 + B4) – (ß1 + B3 + Bs)] [S1]²S2[(B1 + B3 + Bs) + A (82 + B4)] (a2+a4) [S1]+S2[(B1 + B3 + B3)+ A (B2 + B4)] (a1 +a3 + as) [S1]+2 [S1] [(B1 + B3 + Bs) + A (B2 + B4)]8S2 [S1] [(B1 + B3 + Bs) + A (82 + B4)]S+S4 [S1] [(B1 + B3 + Bg) (a2 + a4) + A (B2 + B4) (a1 + az + a5)]S4 [S1] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)]S2[(a1 + a3 + as) – (a2 + a4)][(B1 + B3 + B5) + A (ß2 + B4)]8S2[(a1 + az + a5) – (a2 + a4)][(B1 + B3 + B3) + A (B2 + B4)]8S= 0.where2 [(B1 + B3 + B5)+A (B2 + B4)] ×[(B1 + B3 + B5) (S, + 6) + (B2 + B4) (S1 – 8)]S=-15

The equation given to us is as follows: y1-P=AQ+α1+α3+α5P+α2+α4Qβ1+β3+β5P+β2+β4Q-P When the RHS is…

Answered: The main focus of this article is to…

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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Related questions

Question

Show me the steps of determine green and inf is here

(a1 + a3 + a5) P+(a2+a4) Q
Ул — Р 3D AQ +
- P
(B1 + B3 + B5) P + (B2 + B4) Q
[A (B1 + B3 + B6) – (B2 + B4)]PQ + A (B2 + B4) Q² – (B1 + Ba3 + Bs) P²
(B1 + B3 + B3) P + (B2 + B4) Q
(a1 + a3 + a5) P + (@2+a4)Q
(В + Вз + В5) Р + (В2 + Ba) Q
[A(B1+B3+Bs)-(82+B4)][S1][(B1+B3+Bs)(a2+a4)+A(B2+B4) (a1+a3+a5)]
[(B1+B3+Bs)+A(B2+B4)]*[((B2+B4)-(B1+B3+B5))(A+1)]
([(a1+a3+a5)-(a2+a4)]+8
(B1 + B3 + B5) (2(31+B3+Bs)+A(B2+Ba)]
+ (B2 + B4) (Citag+a5)-(a9+as)]-6
2((B1+B3+B5)+A(B2+B4)]
2
2
A (B2 + B4) (
(@i+a3+as)-(az+as)]-6
2[(B1+B3+B5)+A(B2+B4)]
-(81 + B3 + Bs)
[(@1+a3+as)-(a2+as)]+8
2(B1+B3+Bs)+A(ß2+B4)]
PI T P3 + P5) (231+33+85)+A(62+8) + (B2 + B4) (eIta3+as)-(a2+a4)]-8
[(ai+a3+as)-(a2+a4)]-8
2[(B1+B3+B5)+A(B82+B4)]
I(a1+a3+as)-(a2+a4)]-8
2((B1+33+B5)+A(B2+B4)
(4.24)
[(ai+a3+a5)-(a2+a4)]+8
2[(B1+B3+Bs)+A(B2+B4)]
(a1+a3+as)-(a2+a4)]+ô
2[(B1+B3+B5)+A(B2+B4)]
(a1+a3+as)-(aztas)]+8
2[(B1+B3+B3)+A(B2+B4)]
(a1 + a3 + a5)
+ (a2 + a4)
(B1 + B3 + B5)
+ (B2 + B4)
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
Ym+1 =
Aym+
m = 0,1,2, ..,
В1ут-1 + В2ут-2 + Взут-3 + Ваут-4 + Bзут-5
(1.1)
where the coefficients A, a;, Bi E (0, 00), i = 1, .., 5, while the initial condi-
tions 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 =
B4
when a4 =
= a4 =
= 0 and Eq.(1.1) has been studied in [8] in the special case
B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the
= a5 =
= B5
special case when as = B5 = 0.

Multiplying the denominator and numerator of (4.24) by 4 [(B1 + B3 + B5) + A (B2 + B4)]?
we get
4[A(B1+B3+B5)-(B2+B4)][S1][(B1+ß3+B5)(@2+a4)+A(B2+B4) (a1+a3+æ5)]
[(B2+B4)-(B1+B3+Bs))(A+1)]
Y1 – P =
A (B2 + B4) (S1 – 8)² – (B1 + B3 + Bs) (Sı + 8)²
S
2[(B1 + B3 + B5) + A (B2 + B4)] (aı + a3 + a5) (S1 + 8)
+
S
2[(B1 + B3 + B5) + A (B2 + B4)] (a2 + a4) (S1 – 8)
+
S
4[A(B1+B3+B5)-(B2+Ba)][S1][(B1+B3+Bs)(a2+a4)+A(B2+B4) (a1+a3+a5)]
[((32+B4)-(B1+ß3+B5))(A+1)]
A (B2 + Ba) [S1]? - (B1 + B3 + B5) [S1]?
[A (B2 + B4) – (ß1 + B3 + B3)]82
S
14
2[(B1 + B3 + B5) + A (B2 + B4)] (a2+a4) [S1]
S
2[(B1 + B3 + B5)+ A(B2 + B4)] (a1 + a3 + a5) [S1]
S
2A (B2 + B4) [(a1+az + a5) – (a2 + a4)] & + 2 (B1 + B3 + B5) [S1] &
S
2[(B1 + B3 + B5) + A (B2 + Ba)] (a1+ a3 + a5) 8 – 2[(B1 + B3 + B5) + A (B2 + Ba)] (a2 + a4) 8
S
4[A(B1+B3+B5)-(B2+B4)][S1][(B1+B3+B5)(a2+a4)+A(B2+B4)(a1+a3+as)]
[((B2+B4)-(B1+B3+B5))(A+1)]
S
4[Si][(B1+B3+ßs)(a2+as)+A(ß2+ß4)(@1+a3+as)][A(B2+Ba)-(B1+B3+ßs)1
[((B2+B4)-(B1+ß3+B5))(A+1)]
A (B2 + B4) [S1]? – (B1 + B3 + B3) [S1]²
-
S
[A (B2 + B4) – (ß1 + B3 + Bs)] [S1]²
S
2[(B1 + B3 + Bs) + A (82 + B4)] (a2+a4) [S1]
+
S
2[(B1 + B3 + B3)+ A (B2 + B4)] (a1 +a3 + as) [S1]
+
2 [S1] [(B1 + B3 + Bs) + A (B2 + B4)]8
S
2 [S1] [(B1 + B3 + Bs) + A (82 + B4)]S
+
S
4 [S1] [(B1 + B3 + Bg) (a2 + a4) + A (B2 + B4) (a1 + az + a5)]
S
4 [S1] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)]
S
2[(a1 + a3 + as) – (a2 + a4)][(B1 + B3 + B5) + A (ß2 + B4)]8
S
2[(a1 + az + a5) – (a2 + a4)][(B1 + B3 + B3) + A (B2 + B4)]8
S
= 0.
where
2 [(B1 + B3 + B5)+A (B2 + B4)] ×
[(B1 + B3 + B5) (S, + 6) + (B2 + B4) (S1 – 8)]
S
=
-
15

Expert Solution

Step 1

The equation given to us is as follows:

$y_{1} - P = A Q + \frac{(α_{1} + α_{3} + α_{5}) P + (α_{2} + α_{4}) Q}{(β_{1} + β_{3} + β_{5}) P + (β_{2} + β_{4}) Q} - P$

When the RHS is simplified we will get the equation as follows:

$y_{1} - P = \frac{[A (β_{1} + β_{3} + β_{5}) - (β_{2} + β_{4})] P Q + A (β_{2} + β_{4}) Q^{2} - (β_{1} + β_{3} + β_{5}) P^{2}}{(β_{1} + β_{3} + β_{5}) P + (β_{2} + β_{4}) Q} + \frac{(α_{1} + α_{3} + α_{5}) P + (α_{2} + α_{4}) Q}{(β_{1} + β_{3} + β_{5}) P + (β_{2} + β_{4}) Q} y_{1} - P = \frac{\frac{[A (β_{1} + β_{3} + β_{5}) - (β_{2} + β_{4})] [S_{1}] [(β_{1} + β_{3} + β_{5}) (α_{2} + α_{4}) + A (β_{2} + β_{4}) (α_{1} + α_{3} + α_{5})]}{{[(β_{1} + β_{3} + β_{5}) + A (β_{2} + β_{4})]}^{2} [((β_{2} + β_{4}) - (β_{1} + β_{3} + β_{5})) (A + 1)]}}{(β_{1} + β_{3} + β_{5}) (\frac{[(α_{1} + α_{3} + α_{5}) - (α_{2} + α_{4})] + δ}{2 [(β_{1} + β_{3} + β_{5}) + A (β_{2} + β_{4})]}) + (β_{2} + β_{4}) (\frac{[(α_{1} + α_{3} + α_{5}) - (α_{2} + α_{4})] - δ}{2 [(β_{1} + β_{3} + β_{5}) + A (β_{2} + β_{4})]})} + \frac{A (β_{2} + β_{4}) {(\frac{[(α_{1} + α_{3} + α_{5}) - (α_{2} + α_{4})] - δ}{2 [(β_{1} + β_{3} + β_{5}) + A (β_{2} + β_{4})]})}^{2} - (β_{1} + β_{3} + β_{5}) {(\frac{[(α_{1} + α_{3} + α_{5}) - (α_{2} + α_{4})] + δ}{2 [(β_{1} + β_{3} + β_{5}) + A (β_{2} + β_{4})]})}^{2}}{(β_{1} + β_{3} + β_{5}) (\frac{[(α_{1} + α_{3} + α_{5}) - (α_{2} + α_{4})] + δ}{2 [(β_{1} + β_{3} + β_{5}) + A (β_{2} + β_{4})]}) + (β_{2} + β_{4}) (\frac{[(α_{1} + α_{3} + α_{5}) - (α_{2} + α_{4})] - δ}{2 [(β_{1} + β_{3} + β_{5}) + A (β_{2} + β_{4})]})} + \frac{(α_{1} + α_{3} + α_{5}) (\frac{[(α_{1} + α_{3} + α_{5}) - (α_{2} + α_{4})] + δ}{2 [(β_{1} + β_{3} + β_{5}) + A (β_{2} + β_{4})]}) + (α_{2} + α_{4}) (\frac{[(α_{1} + α_{3} + α_{5}) - (α_{2} + α_{4})] - δ}{2 [(β_{1} + β_{3} + β_{5}) + A (β_{2} + β_{4})]})}{(β_{1} + β_{3} + β_{5}) (\frac{[(α_{1} + α_{3} + α_{5}) - (α_{2} + α_{4})] + δ}{2 [(β_{1} + β_{3} + β_{5}) + A (β_{2} + β_{4})]}) + (β_{2} + β_{4}) (\frac{[(α_{1} + α_{3} + α_{5}) - (α_{2} + α_{4})] - δ}{2 [(β_{1} + β_{3} + β_{5}) + A (β_{2} + β_{4})]})}$

Here $(α_{1} + α_{3} + α_{5}) - (α_{2} + α_{4}) = S_{1}$ .

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