Question

Consider the spring model x″−8x+2x3=0, x ″ − 8 x + 2 x 3 = 0 , we looked at in the previous problem. Linearize the first-order system that you obtained there at the third of the critical points you found. [x′y′]=A[xy] [ x ′ y ′ ] = A [ x y ] , where

Consider the spring model x"-8x2x30, we looked at in the previous problem. Linearize the first-order system that you obtained there at the third of the critical points you found. = A , , where A = Then solve this to find the eigenvalues of the linearized system (enter any complex numbers you may obtain by using "for V-. For real answers, enter them in ascending order; for complex, enter the a - ib root before a + ib): The critical point of the linearized system is A. unstable B. asymptotically stable C. stable and is a(n) A. node B. center C. spiral source D. saddle point E. spiral sink

Answer 1

Given The spring model equation is x'' - 8 x + 2 x^3 = 0 let y = x then y' = x'' = 8x - 2 x^3 = 2 (4x - x^3) Therefore, The system of first order equation is x' = y y' = 2 (4x - x^3) The critical points are given by x' = 0 and y = 0 y = 0 2(4 x - x^3) = 0 4 x - x^3 = 0 x = 0 x = plusorminus 2 Therefore, the critical point FS (negative 2, 0) So, We need to Linear the system at (negative 2, 0) \r\n Let F(x, y) = y and G (x, y) = 8x - 2 x^3 F_x (x, y) = 0 F + y (x, y) = 1 G_x (x, y) = 8 - 6 x^2 G y (x, y) = 0 The Jacobian matrix is I = [Fx (x, y) F_y (x, y) G_x (x, y) G_y (x, y) I = (0 8 - 6x^2 1 0)