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Chapter 4 Solutions
Numerical Analysis
- Using the method of sections need help solving this please explain im stuckarrow_forwardFind all the solutions of the congruence 7x² + 15x = 4 (mod 111).arrow_forward) The set {1,2,..., 22} is to be split into two disjoint non-empty sets S and T in such a way that: (i) the product (mod 23) of any two elements of S lies in S; (ii) the product (mod 23) of any two elements of T lies in S; (iii) the product (mod 23) of any element of S and any element of T lies in T. Prove that the only solution is S = {1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18}, T= {5, 7, 10, 11, 14, 15, 17, 19, 20, 21, 22}.arrow_forward
- 19) Consider this initial value problem: y' + y = 2y = -21² + 2t+ 14, y(0) = 0, y (0) = 0 - What is the solution of the initial value problem?arrow_forward4) Consider the initial value problem " 8y +30y+25y = 0, y(0) = -2, y (0) = 8 What is the t-coordinate of the local extreme value of y = y(t) on the interval (0, ∞)? Enter your answer as a decimal accurate to three decimal places.arrow_forwardTips S ps L 50. lim x2 - 4 x-2x+2 51. lim 22 - X 52. 53. x 0 Answer lim x 0 lim 2-5 X 2x2 2 x² Answer -> 54. lim T - 3x - - 25 +5 b+1 b3b+3 55. lim X x-1 x 1 Answer 56. lim x+2 x 2 x 2 57. lim x²-x-6 x-2 x²+x-2 Answer-> 23-8 58. lim 2-22-2arrow_forward
- 10) Which of the following is the general solution of the homogeneous second-order differential equation y + 8y + 52y=0? Here, C, C₁, and C2 are arbitrary real constants. A) y = C₁ecos(61) + C₂e*sin(61) + C B) y = et (sin(4t) + cos(6t)) + C C) y = C₁esin(6) + C₂e+ cos(6t) + C D) y = C₁esin(6) + C₂e+cos(6) E) y=e(C₁sin(61) + C₂cos(61))arrow_forward3) Consider the initial value problem ' y' + 8y = 0, y(0) = -4, y (0) = 16 What is the solution of this initial value problem? A) y = -4t - 2e8t D) y = -4 + 2e-8t B) y = -2 + 2e8t C) y = -2 -2e-8t E) y = -4+ 2e8t F) y = -2t-2e-8tarrow_forward6) Consider the initial value problem y + cos πι + e²бty = 0, y(-1) = 0, y' (-1) = 0 Which of these statements are true? Select all that apply. A) There exists a nonzero real number r such that y(t) = ert is a solution of the initial value problem. B) The constant function y(t) = -1 is a solution of this initial value problem for all real numbers t. C) The constant function y(t) = 0 is the unique solution of this initial value problem on the interval (-∞, ∞). D) This initial value problem has only one solution on the interval (-7, 5). E) There must exist a function y = q(t) that satisfies this initial value problem on the interval (-7,∞).arrow_forward
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