Question

click here pt) As an illustration of the difficulties that may arise in using the method of undetermined coefficients, consider 2t - -1 + 0 a. Form the complementary solution to the homogeneous equation. ic (t) c1 +c2 is a constant vector, does not work. In fact, if g/p e2a, where a b. Show that seeking a particular solution of the form p(t) a2 had this form, we would arrive at the following contradiction: and

Answer 1

(a) eigen values and eigen vectors of [-1 - 1 -1 -1] lambda = -2, 0 [1 1],[-1 1] Y_c (t) = c_1 e^-2t [1 1] + c_2 e^0t [-1 1] (b) y_p (t) = ae^-2t y_p^t = -2ae^-2t y_p^t = Ay_p + [1 0]e^-2t -2ae^-2t = Aae^-2t + [1 0] (A + 2I) a = [-1 0] [1 -1 -1 1] [a_1 a_2] = [-1 0] a_1 - a_2 = -1: -a_1 + a_2 = 0: a_2 = a_1 a_1 - a_2 = -1: 0 = -1