Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
Let w2n-1 = Un, W2n-2 = Vn, 22n-1 = Xn and z2n-2 = Yn, implies that Vn+1 – ko Un – ko Vn = µ, (11) Un+1 – kį Vn+1 – kị Un = µ, and Yn+1 – Vo Xn – v Yn = €, (12) Xn+1 – Vị Yn+1 - v¼ Xn = €. It is observe that, the equations (11) and (12) are system of linear non-homogeneous difference equations with constants coefficients. First, we solve the system in (11). Thus, consider (11) is (E – ko) Vn – ko Un (13) (E – k1) Un – ki E V, = µ, (14) where E = A + 1, is the shift operator. Multiplying (13) and (14) by (E – kı) and ko, respectively, (E – ki) (E – ko) Vn – (E – k1) ko Un (E – ki) µ, (15) ko (E – k1) Un – ko ki E Vn ko H. (16) %3D Thus, the combination of (15) and (16) is (E² – (ko + ki + ko ki) E + ko ki) Vn = (1– kı + ko) µ, (E² – O E + ko k1) Vn = (1 – k1 + ko) H, (17) where o = ko + kı + ko k1. Clearly, (17) is linear difference equations from order two and it is easy to obtain the solution, $ + V62 – 4 ko ki $ - Vo2 – 4 ko ki 1- k1 + ko + 1- ko - k1 Vn = A + B (18) 2 since A and B are constants. By Substituting (18) in (13), we give V 6² – 4 ko ki 62 – 4 ko ki Un = A ko - 1 0 - Vø2 – 4 ko k1 ko 1 4 ko ki (19) + B - 1 • (G-) (는)-1)~ 1 - ki + ko -- ko – k , H. ko

Let w2n-1 = Un, W2n-2 = Vn, 22n-1 = Xn and z2n-2 = Yn, implies that Vn+1 – ko Un – ko Vn = µ, (11) Un+1 – kį Vn+1 – kị Un = µ, and Yn+1 – Vo Xn – v Yn = €, (12) Xn+1 – Vị Yn+1 - v¼ Xn = €. It is observe that, the equations (11) and (12) are system of linear non-homogeneous difference equations with constants coefficients. First, we solve the system in (11). Thus, consider (11) is (E – ko) Vn – ko Un (13) (E – k1) Un – ki E V, = µ, (14) where E = A + 1, is the shift operator. Multiplying (13) and (14) by (E – kı) and ko, respectively, (E – ki) (E – ko) Vn – (E – k1) ko Un (E – ki) µ, (15) ko (E – k1) Un – ko ki E Vn ko H. (16) %3D Thus, the combination of (15) and (16) is (E² – (ko + ki + ko ki) E + ko ki) Vn = (1– kı + ko) µ, (E² – O E + ko k1) Vn = (1 – k1 + ko) H, (17) where o = ko + kı + ko k1. Clearly, (17) is linear difference equations from order two and it is easy to obtain the solution, $ + V62 – 4 ko ki $ - Vo2 – 4 ko ki 1- k1 + ko + 1- ko - k1 Vn = A + B (18) 2 since A and B are constants. By Substituting (18) in (13), we give V 6² – 4 ko ki 62 – 4 ko ki Un = A ko - 1 0 - Vø2 – 4 ko k1 ko 1 4 ko ki (19) + B - 1 • (G-) (는)-1)~ 1 - ki + ko -- ko – k , H. ko

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
See similar textbooks
icon
Related questions
Question

show me the steps of determine red and all information is here

 
Let w2n-1 = Un, W2n-2 = Vn, %2n–1 = : Xn and z2n-2 = Yn, implies that Vn+1 – ko Un – ko Vn = µ, (11) Un+1 – kį Vn+1 – kį Un = µ, and Yn+1 - Vo Xn – vo Yn = €, (12) Xn+1 vị Yn+1 – Vị Xn = €. It is observe that, the equations (11) and (12) are system of linear non-homogeneous difference equations with constants coefficients. First, we solve the system in (11). Thus, consider (11) is (E – ko) Vn – ko Un (E – ki) Un – ki E V, (13) (14) where E = A + 1, is the shift operator. Multiplying (13) and (14) by (E – k1) and ko, respectively, (E – ki) (E – ko) Vn – (E – k1) ko U, (E – ki) µ, (15) ko (E – k1) Un – ko ki E V, ko fl. (16) Thus, the combination of (15) and (16) is (E² – (ko + ki + ko k1) E + ko ki) Vn = (1– k1 + ko) µ, (E² – ø E + ko k'ì) Vn = (1 – k1 + ko) µ, (17) where o = ko + k1 + ko k1. Clearly, (17) is linear difference equations from order two and it is easy to obtain the solution, 1 – kị + ko 1 – ko – k1 ) 4 ko k1 - 4 ko ki V = A + B (18) 2 since A and B are constants. By Substituting (18) in (13), we give 6? – 4 ko ki o² – 4 ko k1 1 Un = A ko - 1 2 2 1 + B ko – 4 ko k'i 4 ko ki - 1 (19) 1- kı + ko - ko – ki 1 H. + 6
Transcribed Image Text:Let w2n-1 = Un, W2n-2 = Vn, %2n–1 = : Xn and z2n-2 = Yn, implies that Vn+1 – ko Un – ko Vn = µ, (11) Un+1 – kį Vn+1 – kį Un = µ, and Yn+1 - Vo Xn – vo Yn = €, (12) Xn+1 vị Yn+1 – Vị Xn = €. It is observe that, the equations (11) and (12) are system of linear non-homogeneous difference equations with constants coefficients. First, we solve the system in (11). Thus, consider (11) is (E – ko) Vn – ko Un (E – ki) Un – ki E V, (13) (14) where E = A + 1, is the shift operator. Multiplying (13) and (14) by (E – k1) and ko, respectively, (E – ki) (E – ko) Vn – (E – k1) ko U, (E – ki) µ, (15) ko (E – k1) Un – ko ki E V, ko fl. (16) Thus, the combination of (15) and (16) is (E² – (ko + ki + ko k1) E + ko ki) Vn = (1– k1 + ko) µ, (E² – ø E + ko k'ì) Vn = (1 – k1 + ko) µ, (17) where o = ko + k1 + ko k1. Clearly, (17) is linear difference equations from order two and it is easy to obtain the solution, 1 – kị + ko 1 – ko – k1 ) 4 ko k1 - 4 ko ki V = A + B (18) 2 since A and B are constants. By Substituting (18) in (13), we give 6? – 4 ko ki o² – 4 ko k1 1 Un = A ko - 1 2 2 1 + B ko – 4 ko k'i 4 ko ki - 1 (19) 1- kı + ko - ko – ki 1 H. + 6
In this paper, we solve and study the properties of the following system Wn-p 2 Zn-h p=0 Wn-p h=0 E zn-h h=1 p=1 Wn+1 +u and zn+1 = +e, (4) Zn - € Wn - 4 where u and e are arbitrary positive real numbers with initial conditions w; and z; for i = -2, –1,0. Theorem 2.1. Let {wn, z„}-2 be a solution of (4), then 6+ V – 4 ko k1 A - V2 – 4 ko ki 1- ki + ko + w2n-2 +B Ty – Oy - - V2 - 4 ko ki -4 ko ki + V2 – 4 ko ki V2 - 4 ko ki A ko + B ko w2n-1 - 1 2 2 •((금-1) (는) 1– ko + k1 + V2 - 4 vo V1 - V2 - 4 VI 1- v + vo 22n-2 +D %3D 1 + Vu2 – 4 o VI gb + V2 – 4 vo vỊ 1 1. +D - V2 – 4 v vỊ V2 – 4 vo vi 22n-1 - v + vo 1- 0 - VI where A, B, C and D are constants defined as wo -H 2-1 +z-2 ko ki = 0 = ko + ki + ko ki, w-1 + w-2 20 - e 20 - e w-1 +w-2 , = vo + vi + vo V1, 2-1 +2-2 1-Om Vo2 – 4 ko k1 -)") (v - 4ko ki – 4) + 2 (wo - (G ) -). 2 62 - 8 ko ki V2- 4 ko ki 262 - 8 ko ki 1- ko + ki - ko – k1 (1 – kj + ko 1- k1 - ko B = u- w-2 +2 wn- - vi +u C = -2 + 2 22 - 8 0 vI - o - VI ) [(는 1- v1 + o - v0 - v1 1- v + o 2) (V2 – 4vo vI + +2 2 2 -- 0 - v1 since ko + ki #1 and vo + vı #1 for n e N. Proof. To obtain the expressions of the general solutions for (4), we rewrite it in the follow form 2 zn-h Wn-p p=1 Wn+1 h=1 Zn+1 and (5) 1 E Wn-p 1 > Zn-h h=0 wn -u p-0 4 Then, we assume that Wn+1 - l Wn - H kn+1 = 1 kn = 2 Wn-p p=0 Wn-p p=1 (6) Zn+1 -€ Zn - € Vn+1 = Vn 2 E Zn-h > Zn-h h=1 h=0 Substituting (6) in (5), we have Zn-h h=1 1 == kn-1, Vn kn+1 = (7) E Wn-p p=1 Vn+1 = = Vn-1. kn %3D Wn - Hence, we see that 1 ki = -, Vo wo - H 20 - € ko w-1+ w_2 Vo = Z-1+ z-2 ko and at n = 1, k2 = ko and V2 = Vo, at n = 2, k3 = k1 and V3 = V1, at n = 3, k4 = ko and V4 = Vo, at n = 4, ks = k and V5 = V1, (8) at n = 2 n, k2n = ko and V2n = Vo, at n = 2n + 1, k2n+1 = k, and V2n+1 = V1. Now, from the relations in (6) we get Wn = µ+ kn (wn-1 + wn-2) and zn = e + vn (žn-1 + zn-2). (9) Using (8) in (9), we have W2n = H+ ko (w2n-1 + w2n-2), W2n+1 = µ+ k1 (w2n + w2n-1), (10) Z2n = €+ vo (z2n-1 + 2n-2), 22n+1 = €+ V1 (22n + 22n-1).
Transcribed Image Text:In this paper, we solve and study the properties of the following system Wn-p 2 Zn-h p=0 Wn-p h=0 E zn-h h=1 p=1 Wn+1 +u and zn+1 = +e, (4) Zn - € Wn - 4 where u and e are arbitrary positive real numbers with initial conditions w; and z; for i = -2, –1,0. Theorem 2.1. Let {wn, z„}-2 be a solution of (4), then 6+ V – 4 ko k1 A - V2 – 4 ko ki 1- ki + ko + w2n-2 +B Ty – Oy - - V2 - 4 ko ki -4 ko ki + V2 – 4 ko ki V2 - 4 ko ki A ko + B ko w2n-1 - 1 2 2 •((금-1) (는) 1– ko + k1 + V2 - 4 vo V1 - V2 - 4 VI 1- v + vo 22n-2 +D %3D 1 + Vu2 – 4 o VI gb + V2 – 4 vo vỊ 1 1. +D - V2 – 4 v vỊ V2 – 4 vo vi 22n-1 - v + vo 1- 0 - VI where A, B, C and D are constants defined as wo -H 2-1 +z-2 ko ki = 0 = ko + ki + ko ki, w-1 + w-2 20 - e 20 - e w-1 +w-2 , = vo + vi + vo V1, 2-1 +2-2 1-Om Vo2 – 4 ko k1 -)") (v - 4ko ki – 4) + 2 (wo - (G ) -). 2 62 - 8 ko ki V2- 4 ko ki 262 - 8 ko ki 1- ko + ki - ko – k1 (1 – kj + ko 1- k1 - ko B = u- w-2 +2 wn- - vi +u C = -2 + 2 22 - 8 0 vI - o - VI ) [(는 1- v1 + o - v0 - v1 1- v + o 2) (V2 – 4vo vI + +2 2 2 -- 0 - v1 since ko + ki #1 and vo + vı #1 for n e N. Proof. To obtain the expressions of the general solutions for (4), we rewrite it in the follow form 2 zn-h Wn-p p=1 Wn+1 h=1 Zn+1 and (5) 1 E Wn-p 1 > Zn-h h=0 wn -u p-0 4 Then, we assume that Wn+1 - l Wn - H kn+1 = 1 kn = 2 Wn-p p=0 Wn-p p=1 (6) Zn+1 -€ Zn - € Vn+1 = Vn 2 E Zn-h > Zn-h h=1 h=0 Substituting (6) in (5), we have Zn-h h=1 1 == kn-1, Vn kn+1 = (7) E Wn-p p=1 Vn+1 = = Vn-1. kn %3D Wn - Hence, we see that 1 ki = -, Vo wo - H 20 - € ko w-1+ w_2 Vo = Z-1+ z-2 ko and at n = 1, k2 = ko and V2 = Vo, at n = 2, k3 = k1 and V3 = V1, at n = 3, k4 = ko and V4 = Vo, at n = 4, ks = k and V5 = V1, (8) at n = 2 n, k2n = ko and V2n = Vo, at n = 2n + 1, k2n+1 = k, and V2n+1 = V1. Now, from the relations in (6) we get Wn = µ+ kn (wn-1 + wn-2) and zn = e + vn (žn-1 + zn-2). (9) Using (8) in (9), we have W2n = H+ ko (w2n-1 + w2n-2), W2n+1 = µ+ k1 (w2n + w2n-1), (10) Z2n = €+ vo (z2n-1 + 2n-2), 22n+1 = €+ V1 (22n + 22n-1).
Expert Solution
steps

Step by step

Solved in 5 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,
SEE MORE TEXTBOOKS