= a, duo b (e uz+g uz+s u4) – d (c uz+f u3+r us) (d ui+e uz+g uz+su4)² aF(u0,...,us) %3D Ine ƏF(uo,...,us) duz c (d ui+g uz+s u4) - e (b ui+f u3+r u4) (d u1+e uz+g u3+sus)? ƏF(uo,.u4) f (d u1+e uz2+s u4) – g (b ui+c uz+r u4) duz (d ui+e uz+g u3+su4)² ƏF(uo....,u4) dus r (d ui+e uz+g u3) – s (b u1+c uz+f u3) (d ui+e uz+g u3+su4)?

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Show me the steps of determine red and it complete

2
The local stability of the solutions
In this section, we study the local stability of the solutions of Eq.(1). The
equilibrium point a of Eq.(1) is the positive solution of the equation
b +c+f +r
d +e +g+ s'
x = ax +
(6)
If a < 1, then the only positive equilibrium point z of Eq.(1) is given by
b +c+f + r
(7)
(1 – a) ( d+ e + g+s)
Let us now introduce a continuous function F : (0, 00)5 → (0, 00) which
is defined by
buj + cu2 + fu3 + ru4
F(uo,.., u4) = auo +
(8)
dui + eu2 + gu3 + su4
Therefore it follows that
OF (uo,...,u4)
= a,
aF(uo,..,u4)
b (e uz+g uz+8 u4) – d (c u2+f u3+r us)
(d ui+e u2+g u3+su4)2
OF(uo,...,u4)
au2
e (d ui+g uz+s u4) – e (b u1+f u3+r us)
(d u1+e uz+g uz+su4)²
OF (uo,...,u4)
duz
f (d ui+e uz+s u4) – g (b ui+c u2+r us)
(d u1+e u2+g uz+su4)?
aF(uo,...,u4)
r (d ui+e u2+g u3) – s (b u1+c u2+f u3)
(d u1+e u2+g_u3+su4)²
Consequently, we get
OF (ã,...,ã)
Juo
= a = - P4:
aF(7,...,ï)
(1-a) [b (e +g +s) – d (c +f +r)]
(d +e +g+s)(b +c +f+r)
= - P3,
OF (7,..,ã)
(1-a) [e (d +g +s) – e (b +f +r)]
(d +e +g+s)(b +c +f+r)
= - P2,
(9)
(1-a)[f (d +e +s) – g (b +c+r)]
(d +e +g+s)(b +c +f+r)
= - P1,
aF (7,...,ã)
(1-a)[r (d +e +g) - 8 (b +c+f)]
(d +e +g+s)(b +c +f+r)
= - Po.
Thus, the linearized equation of Eq.(1) about a takes the form
Yn+1 + P4Yn + P3Yn-1+ P2Yn-2 + P1Yn-3 + Poyn-4 = 0,
(10)
Transcribed Image Text:2 The local stability of the solutions In this section, we study the local stability of the solutions of Eq.(1). The equilibrium point a of Eq.(1) is the positive solution of the equation b +c+f +r d +e +g+ s' x = ax + (6) If a < 1, then the only positive equilibrium point z of Eq.(1) is given by b +c+f + r (7) (1 – a) ( d+ e + g+s) Let us now introduce a continuous function F : (0, 00)5 → (0, 00) which is defined by buj + cu2 + fu3 + ru4 F(uo,.., u4) = auo + (8) dui + eu2 + gu3 + su4 Therefore it follows that OF (uo,...,u4) = a, aF(uo,..,u4) b (e uz+g uz+8 u4) – d (c u2+f u3+r us) (d ui+e u2+g u3+su4)2 OF(uo,...,u4) au2 e (d ui+g uz+s u4) – e (b u1+f u3+r us) (d u1+e uz+g uz+su4)² OF (uo,...,u4) duz f (d ui+e uz+s u4) – g (b ui+c u2+r us) (d u1+e u2+g uz+su4)? aF(uo,...,u4) r (d ui+e u2+g u3) – s (b u1+c u2+f u3) (d u1+e u2+g_u3+su4)² Consequently, we get OF (ã,...,ã) Juo = a = - P4: aF(7,...,ï) (1-a) [b (e +g +s) – d (c +f +r)] (d +e +g+s)(b +c +f+r) = - P3, OF (7,..,ã) (1-a) [e (d +g +s) – e (b +f +r)] (d +e +g+s)(b +c +f+r) = - P2, (9) (1-a)[f (d +e +s) – g (b +c+r)] (d +e +g+s)(b +c +f+r) = - P1, aF (7,...,ã) (1-a)[r (d +e +g) - 8 (b +c+f)] (d +e +g+s)(b +c +f+r) = - Po. Thus, the linearized equation of Eq.(1) about a takes the form Yn+1 + P4Yn + P3Yn-1+ P2Yn-2 + P1Yn-3 + Poyn-4 = 0, (10)
l stc ksa
12:14 AM
@ 1 60% 4
The objective of this article is to investigate some qualitative behavior of
the solutions of the nonlinear difference equation
bxn-1 + cæn-2+ fxn-3 + ræn-4
Xn+1 = axn +
n = 0, 1, 2, . (1)
dxn-1+ exn-2 + gæn-3 + sxn-4
where the coefficients a, b, c, d, e, f, g, r, s E (0, 00), while the initial con-
ditions a_4,x_3,x_2, x-1, xo are arbitrary positive real numbers. Note that
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Transcribed Image Text:l stc ksa 12:14 AM @ 1 60% 4 The objective of this article is to investigate some qualitative behavior of the solutions of the nonlinear difference equation bxn-1 + cæn-2+ fxn-3 + ræn-4 Xn+1 = axn + n = 0, 1, 2, . (1) dxn-1+ exn-2 + gæn-3 + sxn-4 where the coefficients a, b, c, d, e, f, g, r, s E (0, 00), while the initial con- ditions a_4,x_3,x_2, x-1, xo are arbitrary positive real numbers. Note that Cancel Actual Size (399 KB) Choose
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