Solve the initial value problem: y\" - 4y\' + 3y = 10e^3t with y(0) = 0 and y\'(

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Solve the initial value problem: y" - 4y' + 3y = 10e^3t with y(0) = 0 and y'(0) = 0. For the homogeneous equation the characteristic equation is r^2 - 4r + 3 (r - 3)(r - 1) = 0 and so y_h = C_1^e^3t + C_2^e^t. Y_p^1 = Ae^3t and since the associated root, 3 is a a root of the homogeneous, Y_p = Ate^3t. 3 times Y_p = Ate^3t -4 times Y_p^1 = 3 Ate^3t + Ae^3t 1 times Y"_p/10e^3t = 9 Ate^3t + 6Ae^3t/0 + 2Ae^3t So A = 5, and y_g = C_1^e^3t + C_2^e^t + 5te^3t, and y'_g = (3C_1 + 5)e^3t + C_2^e^t + 15te^3t. 0 = y(0) = C_1 + C_2, 0 = y'(0) = 3C_1 + 5 + C_2. So -C_2 = C_1 = -5/2. y = 5/2e^3t + 5/2 e^t + 5te^3t.

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