Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
Uzn – V2n = (40 – vo)J2n+1 – (u–1 – v_1)2n+2+b(u_2– v-2) J2n for n>-1. From (2.35) uzn + v2n = (40+ vo) J2n+1+(u-1+v-1) J2n+2+b(u_2+v-2) J2n, (2.37) (2.38) for n>-1. By summing the equations (2.37) and (2.38) we get (J2n+1 +J2n+1) uo+ (J2n+1 – J½n+1) vo+ (J2n+2– J2n+2) u-1 | U2n 2 (J2n+2 +Jn+2) v-1 +b(J2n+J½n)u_2+b (J2n- J2n) v_2 (2.39) 2 for n>-1. By subtracting equation (2.37) from equation (2.38), we have (J2n+1 - Jn+1) uo+(J2n+1 +J½n+1) Vo + (J2n+2+J%n+2) u-1 V2n = (J2n+2 – Jan+2) v-1 +b(J2n– J2n) u–2 +b (J2n+J½n) v-2 (2.40) for n>-1. From (2.36) we have uzn+1 – V2n+1 = - (uo – vo) Jźn+2 +(u-1 – v-1) J2n+3 - b (u-2 –v-2)J2n+1» (2.41) for n > -1. From (2.35) uzn+1+V2n+1 = (uo+ vo) J2n+2+ (u-i+v-1) J2n+3+b (u-2+v-2)J2n+1, (2.42) for n>-1. By summing the equations (2.41) and (2.42) we get (J2n+2- Jan+2) 4o+ (J2n+2+J½n+2) vo+ (J2n+3+J%n+3) u_1 U2n+1 = (Jan+3- Jn+3) v-1+b (Jzn+1 – Jan+1) u-2 , b(J2n+1+J½n+1) v-2 + (2.43) for n> -1. By subtracting equation (2.41) from equation (2.42), we have

Uzn – V2n = (40 – vo)J2n+1 – (u–1 – v_1)2n+2+b(u_2– v-2) J2n for n>-1. From (2.35) uzn + v2n = (40+ vo) J2n+1+(u-1+v-1) J2n+2+b(u_2+v-2) J2n, (2.37) (2.38) for n>-1. By summing the equations (2.37) and (2.38) we get (J2n+1 +J2n+1) uo+ (J2n+1 – J½n+1) vo+ (J2n+2– J2n+2) u-1 | U2n 2 (J2n+2 +Jn+2) v-1 +b(J2n+J½n)u_2+b (J2n- J2n) v_2 (2.39) 2 for n>-1. By subtracting equation (2.37) from equation (2.38), we have (J2n+1 - Jn+1) uo+(J2n+1 +J½n+1) Vo + (J2n+2+J%n+2) u-1 V2n = (J2n+2 – Jan+2) v-1 +b(J2n– J2n) u–2 +b (J2n+J½n) v-2 (2.40) for n>-1. From (2.36) we have uzn+1 – V2n+1 = - (uo – vo) Jźn+2 +(u-1 – v-1) J2n+3 - b (u-2 –v-2)J2n+1» (2.41) for n > -1. From (2.35) uzn+1+V2n+1 = (uo+ vo) J2n+2+ (u-i+v-1) J2n+3+b (u-2+v-2)J2n+1, (2.42) for n>-1. By summing the equations (2.41) and (2.42) we get (J2n+2- Jan+2) 4o+ (J2n+2+J½n+2) vo+ (J2n+3+J%n+3) u_1 U2n+1 = (Jan+3- Jn+3) v-1+b (Jzn+1 – Jan+1) u-2 , b(J2n+1+J½n+1) v-2 + (2.43) for n> -1. By subtracting equation (2.41) from equation (2.42), we have

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 determine yellow and all information is here

constant coefficients Xn+1 = axn + bxn-1+cxn-2, nE No, (1.4) which has actually the general solution X, = xoSn+x-1 (Sn+1 - aSn) +cx_2Sn-1, nE No, (1.5) where (S.). of equation (1.4) satisfving the initial values S_2 = S_ 0. So= L The equation axn-IXn-k Xn+1 = nE No, (1.1) 6. bxn-p where the initial conditions are arbitrary positive real numbers, k, 1, p, q are non- negative integers and a, b, c are positive constants, is one of the difference equations whose solutions are associated with number sequences. Positive solutions of concrete Motivated by this line of investigations, here we show that the systems of differ- ence equations Xn+l = f(af (Pn-1)+bf (qn-2)), yn+1 = f'(af (rn-1)+bf (Sn-2)), (1.8) for n E No, where the sequences Pn, qn, In and Sn are some of the sequences xn and Yn, f : Df → R is a "1 – 1" continuous function on its domain Df C R, the initial values x-j, y-j, je {0,1,2} are arbitrary real numbers and the parameters and a, b In this section, we consider the eight special cases of systems (1.8), where the sequences Pn, 9n, rn, Sn are some of the sequences x, and yn, for n >-2, and initial values x-j, y-j. j e {0, 1,2}, are arbitrary real numbers. 2.1. Case 1: Pn = Xn, qn = Xn, In = Yn, Sn = yn In this case, system (1.8) is expressed as Xn+1 =f"(af (xn-1)+bf (xn-2)), Yn+1 = f (af (n-1)+bf (Yn-2)), (2.1) for n E No. Since f is "1-1", from (2.1) f (Xn+1) = af (xn-1)+bf (xn-2), f(Vn+1) =af (ya-1)+bf Va-2), (2.2) for n E No. By using the change of variables f (xn): = Un, and f(yn) = Vn, n2 -2, (2.3) system (2.2) is transformed to the following one Un+l = aun-1+bun-2, Vn+l = avn-1+bvn-2, (2.4) for n E No. By taking a = 0, b=a, c= b in (1.4) and S, = Jn+1, for all n >-2, which is called generalized Padovan sequence, in (1.5), the solutions to equations in (2.4) are given by U, = uoJn+1+u-1Jn+2+bu-2Jn, (2.5) Vn = voJn+1+v-1Jn+2+bv_2Jn, (2.6) for n E No. From (2.3), (2.5) and (2.6), it follows that the general solution to system (2.2) is given by Xn =f(F (x0) Jn+1 +f (x-1)/n+2+bf (x-2)J.), n> -2, Ya =f'(S (vo) Jn+1+f(y-1)Ja+2+bf (y-2) Ja), n2-2. (2.7) (2.8) 2.2. Case 2: Pn = Xn, qn = Xn, In =Xn, Sn = Xn In this case, system (1.8) becomes Xa+1 =f"(af (xn-1)+bf (xn-2)), Ynt1 =f"(af (*-1)+bf (xn-2)), (2.9) for n e No. It should be first note that from the equations in (2.9) it immediately follows that x, = yn, for all n E N. From (2.7), the general solution to system (2.9) is X = Ya =f'(f (x0) Jn+1 +f (x-1) Jn+2+bf (x-2)Jn), nɛN. (2.10)
Transcribed Image Text:constant coefficients Xn+1 = axn + bxn-1+cxn-2, nE No, (1.4) which has actually the general solution X, = xoSn+x-1 (Sn+1 - aSn) +cx_2Sn-1, nE No, (1.5) where (S.). of equation (1.4) satisfving the initial values S_2 = S_ 0. So= L The equation axn-IXn-k Xn+1 = nE No, (1.1) 6. bxn-p where the initial conditions are arbitrary positive real numbers, k, 1, p, q are non- negative integers and a, b, c are positive constants, is one of the difference equations whose solutions are associated with number sequences. Positive solutions of concrete Motivated by this line of investigations, here we show that the systems of differ- ence equations Xn+l = f(af (Pn-1)+bf (qn-2)), yn+1 = f'(af (rn-1)+bf (Sn-2)), (1.8) for n E No, where the sequences Pn, qn, In and Sn are some of the sequences xn and Yn, f : Df → R is a "1 – 1" continuous function on its domain Df C R, the initial values x-j, y-j, je {0,1,2} are arbitrary real numbers and the parameters and a, b In this section, we consider the eight special cases of systems (1.8), where the sequences Pn, 9n, rn, Sn are some of the sequences x, and yn, for n >-2, and initial values x-j, y-j. j e {0, 1,2}, are arbitrary real numbers. 2.1. Case 1: Pn = Xn, qn = Xn, In = Yn, Sn = yn In this case, system (1.8) is expressed as Xn+1 =f"(af (xn-1)+bf (xn-2)), Yn+1 = f (af (n-1)+bf (Yn-2)), (2.1) for n E No. Since f is "1-1", from (2.1) f (Xn+1) = af (xn-1)+bf (xn-2), f(Vn+1) =af (ya-1)+bf Va-2), (2.2) for n E No. By using the change of variables f (xn): = Un, and f(yn) = Vn, n2 -2, (2.3) system (2.2) is transformed to the following one Un+l = aun-1+bun-2, Vn+l = avn-1+bvn-2, (2.4) for n E No. By taking a = 0, b=a, c= b in (1.4) and S, = Jn+1, for all n >-2, which is called generalized Padovan sequence, in (1.5), the solutions to equations in (2.4) are given by U, = uoJn+1+u-1Jn+2+bu-2Jn, (2.5) Vn = voJn+1+v-1Jn+2+bv_2Jn, (2.6) for n E No. From (2.3), (2.5) and (2.6), it follows that the general solution to system (2.2) is given by Xn =f(F (x0) Jn+1 +f (x-1)/n+2+bf (x-2)J.), n> -2, Ya =f'(S (vo) Jn+1+f(y-1)Ja+2+bf (y-2) Ja), n2-2. (2.7) (2.8) 2.2. Case 2: Pn = Xn, qn = Xn, In =Xn, Sn = Xn In this case, system (1.8) becomes Xa+1 =f"(af (xn-1)+bf (xn-2)), Ynt1 =f"(af (*-1)+bf (xn-2)), (2.9) for n e No. It should be first note that from the equations in (2.9) it immediately follows that x, = yn, for all n E N. From (2.7), the general solution to system (2.9) is X = Ya =f'(f (x0) Jn+1 +f (x-1) Jn+2+bf (x-2)Jn), nɛN. (2.10)
2.6. Сase 6: р, — Ул, qn 3 Ул, Гл — Хл, Sn — Xn In this case, we obtain the system Xn+1 =f (af (yn-1)+bf (yn-2)), Yn+1 = f'(af (xp-1)+bf(xn-2)), (2.29) %3D for n E No. Since f is "1 – 1", from (2.29) f (Xn+1) = af (yn-1)+bf (yn-2), f(yn+1) = af (xp-1)+bf (xn=2), (2.30) for n e No. By using the change of variables f (xn) = = Un, and f(yn) = Vn, n> -2, (2.31) system (2.30) is transformed to the following one Un+1 = avn-1+bvn-2, Vn+1 = aun-1+ bun-2, n E No. (2.32) By summing the equations in (2.32) we get Un+1+Vn+1 =a (un-1+Vn-1) +b (un-2+Vn-2), n E No, (2.33) whereas by subtracting the second one from the first, we have Un+1 – Vn+l =-a(un-1- Vn-1) - b(u-2- Vn-2), n E No. (2.34) From (2.5), we can write the solution of equation (2.33) as Un + Vn = (uo+ vo) Jn+1 + (u-1+v-1)Jn+2+b (u-2+v-2)Jn, (2.35) for n > -2. On the other hand, by taking a = 0, b = -a, c = -b in (1.4) and S, = (-1)" J+1 for all n > -2, which is called generalized Padovan sequence, in (1.5), from (2.34), we also have Un – Vn = (–1)"( (u0 – vo)Ji+1 – (u-1 – v_1)n+2+b (u-2– v-2) .), (2.36) for n >-2. From (2.36) we obtain Uzn – V2n = (4o – v) 2n+1– (u–1 – v_1)2n+2+b(u_2– v_2) 2n (2.37) for n>-1. From (2.35) uzn + V2n = (uo+ vo) J2n+1+ (u-1 +v-1) J2n+2+b (u_2+v-2)J2n (2.38) for n> -1. By summing the equations (2.37) and (2.38) we get (Jan+1 +J2n+1) 4o+ (J2n+1 - J2n+1) vo+ (J2n+2- J2n+2) u–1 U2n 2 (Jan+2+J½n+2) v-1 +b(J2n+J½n)u_2+b(J2n – J,) v-2 + (2.39) for n>-1. By subtracting equation (2.37) from equation (2.38), we have (Jan+1 - J2n+1) uo + (J2n+1+J½n+1) vo+(J2n+2+J%n+2) u-I V2n = 2 (Jan+2- Jan+2) v-1 +b(J2n– Jn) u–2+b (J2n +J½n) v-2 (2.40) 2 for n> -1. From (2.36) we have – (40 – vo) J2n+2+(u-1 – v-1)J2n+3 – b (u-2 – v-2) J2n+1» (2.41) U2n+1 - V2n+1 - V-1 for n>-1. From (2.35) U2n+1 +V2n+l = (uo + vo) J2n+2+ (u-1 +v-1) J2n+3+b (u-2+v-2) J2n+1, (2.42) for n>-1. By summing the equations (2.41) and (2.42) we get (J2n+2- Jin+2) 4o+ (J2n+2+J½n+2) vo+ (J2n+3+J%n+3) u_1 U2n+1 = 2 (J2n+3- Jn+3) v-1+b (J2n+1 – J½n+1) u-2 , b(J2n+1+J½n+1) v-2 2 2 (2.43) for n>-1. By subtracting equation (2.41) from equation (2.42), we have (J2n+2+J2n+2) uo+(J2n+2– J2n+2) vo+ (J2n+3 – J½n+3) u–1 | V2n+1 = 2
Transcribed Image Text:2.6. Сase 6: р, — Ул, qn 3 Ул, Гл — Хл, Sn — Xn In this case, we obtain the system Xn+1 =f (af (yn-1)+bf (yn-2)), Yn+1 = f'(af (xp-1)+bf(xn-2)), (2.29) %3D for n E No. Since f is "1 – 1", from (2.29) f (Xn+1) = af (yn-1)+bf (yn-2), f(yn+1) = af (xp-1)+bf (xn=2), (2.30) for n e No. By using the change of variables f (xn) = = Un, and f(yn) = Vn, n> -2, (2.31) system (2.30) is transformed to the following one Un+1 = avn-1+bvn-2, Vn+1 = aun-1+ bun-2, n E No. (2.32) By summing the equations in (2.32) we get Un+1+Vn+1 =a (un-1+Vn-1) +b (un-2+Vn-2), n E No, (2.33) whereas by subtracting the second one from the first, we have Un+1 – Vn+l =-a(un-1- Vn-1) - b(u-2- Vn-2), n E No. (2.34) From (2.5), we can write the solution of equation (2.33) as Un + Vn = (uo+ vo) Jn+1 + (u-1+v-1)Jn+2+b (u-2+v-2)Jn, (2.35) for n > -2. On the other hand, by taking a = 0, b = -a, c = -b in (1.4) and S, = (-1)" J+1 for all n > -2, which is called generalized Padovan sequence, in (1.5), from (2.34), we also have Un – Vn = (–1)"( (u0 – vo)Ji+1 – (u-1 – v_1)n+2+b (u-2– v-2) .), (2.36) for n >-2. From (2.36) we obtain Uzn – V2n = (4o – v) 2n+1– (u–1 – v_1)2n+2+b(u_2– v_2) 2n (2.37) for n>-1. From (2.35) uzn + V2n = (uo+ vo) J2n+1+ (u-1 +v-1) J2n+2+b (u_2+v-2)J2n (2.38) for n> -1. By summing the equations (2.37) and (2.38) we get (Jan+1 +J2n+1) 4o+ (J2n+1 - J2n+1) vo+ (J2n+2- J2n+2) u–1 U2n 2 (Jan+2+J½n+2) v-1 +b(J2n+J½n)u_2+b(J2n – J,) v-2 + (2.39) for n>-1. By subtracting equation (2.37) from equation (2.38), we have (Jan+1 - J2n+1) uo + (J2n+1+J½n+1) vo+(J2n+2+J%n+2) u-I V2n = 2 (Jan+2- Jan+2) v-1 +b(J2n– Jn) u–2+b (J2n +J½n) v-2 (2.40) 2 for n> -1. From (2.36) we have – (40 – vo) J2n+2+(u-1 – v-1)J2n+3 – b (u-2 – v-2) J2n+1» (2.41) U2n+1 - V2n+1 - V-1 for n>-1. From (2.35) U2n+1 +V2n+l = (uo + vo) J2n+2+ (u-1 +v-1) J2n+3+b (u-2+v-2) J2n+1, (2.42) for n>-1. By summing the equations (2.41) and (2.42) we get (J2n+2- Jin+2) 4o+ (J2n+2+J½n+2) vo+ (J2n+3+J%n+3) u_1 U2n+1 = 2 (J2n+3- Jn+3) v-1+b (J2n+1 – J½n+1) u-2 , b(J2n+1+J½n+1) v-2 2 2 (2.43) for n>-1. By subtracting equation (2.41) from equation (2.42), we have (J2n+2+J2n+2) uo+(J2n+2– J2n+2) vo+ (J2n+3 – J½n+3) u–1 | V2n+1 = 2
Expert Solution
steps

Step by step

Solved in 3 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