Which of the following recurrence relations is nonlinear ?(1) an - 3an-1 - 50n-2 = 0an - 3n²an-1-5an-2 = 0(2) an(3) an 3an-1 an 2 + 4 = 03an-1-5an-2 +4 = 0(4) anAB-Choice (2)Choice (1)(C) Choice (4)D) Choice (3)

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Answered: Which of the following recurrence…

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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2. Solve these recurrence relations together with the initial conditions given. (а) ап 3 ап-1 + 6а,-2 for n 22, ао 3D 3, ај %3D 6 (b) а, — 7ал-1 — 10а, -2 for n > 2, ag 3D 2, ај 3 1 бал-1 — 8ап-2 for n > 2, ao %3 4, ај %—D 10 (c) an = аn-2 for n > 2, ао — 4, ај —D 1 (е) а, — а,-2 for n > 2, ag 3D 5, ај %3D -1 (f) а, — — 6ал -1 - 9ат-2 for n 22, ao %3D 3, ај 3D -3 (8) а,+2 3— 4a, +1 + 5а, for n 20, ag 3D 2, а, 3 8. (d) an = 2an-1 °
The closed form solution to the recurrence relation a(n) = a(n-1) + 4 with a(0) = 3 is (A) a (n+1)= a(n) - 3 (B) a(n) = 4 + 3n (C) a(n) = 3 + 4n (D) a(n) = 4n-3 (E) None of the above.
What is the general solution for the recurrence relation tn = 13tn-1 – 36tn-2 ? (1) tn = c1(-4)" + c2(9)" (2) tn = c1(4)" + c2(9)" (3) tn = c1(-4)" + c2(-9)" (4) tn = c1(4)" + c2(-9)"
Explain how the telescoping technique produces solutions to recurrence relations. Use a₁ = an-1 + 2n with ao = 5 as an example to elaborate.
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What is the form of the particular solution of the linear non-homogeneous recurrence relation. [hint: the characteristic equation factorised is (x+5)²(x+1)²] an = -12an-1-36an-2 - 60an-3-25an-4+ (3n³ + 1) (-5)" - 1 −
Which of the following are second-order linear homogeneous recurrence relations with constant coefficients? (Select all that apply.) n?k = (k - 1)ak - 1 + 2kag - 2 bk = -bk - 1+ 7bk - 2 Ck = 3Ck - 1 + 1 - 3d2, + d k - 1 k - 2 ロrk=「k-1- 6rk- 3 Sk = Sk - 1 105k - 2 %3D + Need Help? Read It
an = Identify the properties of the given recurrence relations. 3an- 7-1+4an-21 Multiple Choice + 5an-3 nonlinear and homogeneous with constant coefficients and degree 2 linear and homogeneous with constant coefficients and degree 3 linear and nonhomogeneous with constant coefficients and degree 3 linear and homogeneous without constant coefficients and degree 3
6. Solve please
Find the solution to the recurrence relation by using an iterative approach. The recurrence relation a, = -5a, - 1 with the initial condition ao = 3 a. The solution for the recurrence relation is an = -5a, - 1 = (-5)?a,-2 = - .. = (-5)"an - n = (-5)"ao = 3 - (-5)" O b. The solution for the recurrence relation is an = -5an - 1 = 5an - 2 =- · = 5a, - n = 5a0 = 5(3) = 15 %3D O c. The solution for the recurrence relation is an = -5a, - 1 = (-5)²a, - 2 = · . = (-5)"a, - n = (-5)°ao = 1.3 = 3 %3D d. The solution for the recurrence relation is an = -5a, - 1 = -5an -2 = - -5a, - = -5a0 = -5(3) = 15
tion 6 The value of the characteristics variable r in the characteristic ret equation of the recurrence relation an = -2an 1+3an 2 is ered ed out of Or=-1,-3 ag question Or= 1,3 Or= 1,-3 Or=-1,3 evious page Next pag

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Which of the following recurrence relations is nonlinear ? (1) an - 3an-1 - 5an-2 = 0 (2) an - 3n²an-1 – 5an-2 = 0 - (3) an 3an-1-an-2+4= 0 (4) an 3an-1-5an-2 +4 = 0 - A) Choice (2) B Choice (1) (C) Choice (4) D Choice (3)