Question: What is the solution of the initial value problem y + 36y = 0, y (41) = -2, y' (4x) = 5? O 5 y

 What is the solution of the initial value problem y" +36y = 0, y (41) = -2, y' (4x) = 5? O5 y = -2 cos(6t) + -sin(6t) Oy = -2 sin(6t) +

What is the solution of the initial value problem y" + 36y = 0, y (41) = -2, y' (4x) = 5? O 5 y = -2 cos(6t) + -sin(6t) Oy = -2 sin(6t) + 5 cos(6t) O y = cos(6t) + sin(6t) O 5 y = -2e361 + - e -36t 6 Oy = -2e-6 + 5e-6tWhich of the following are solutions to the homogeneous second-order differential equation y" - 10y' + 29y = 0? Select all that apply. O y1 = Cie- sin(2t) + Cze-St cos(2t), where C1 and C2 are any real constants O y2 = Cest cos(2t), where C is any real constant O y3 = 3est (sin(2t) + cos(2t)) O y4 = e-st cos(2t) O ys = es cos(2t) O V6 = - sin(2t) N O y7 = 5estX Good. Solve the initial value problem y" + 12y' + 37y = 0, y(0) = 1, y' (0) = 0 O y = be of cost + e- sint O y = 6e-6 + e-6t O y = cost + 6 sint y = 6cost + sint O y = e-of + 6e-6t O y = e- cost + 6e- sint

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