inverse laplace of (3s)/(s^2-2s+12) Solution By partial fractions, (3s + 2) / [(

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By partial fractions, (3s + 2) / [(s - 2)(s^2 + 2s + 10)] = A/(s - 2) + (Bs + C)/(s^2 + 2s + 10). To solve for A, B, C, first clear the denominators: 3s + 2 = A(s^2 + 2s + 10) + (Bs + C)(s - 2). Let s = 2: 8 = 18A + 0 ==> A = 4/9. So, 3s + 2 = (4/9)(s^2 + 2s + 10) + (Bs + C)(s - 2) ==> 0 = (4s - 11) + 9(Bs + C) ==> 9C - 11 = 0 and 9B + 4 = 0 ==> B = -4/9 and C = 11/9. Therefore, (3s + 2) / [(s - 2)(s^2 + 2s + 10)] = (1/9) [4/(s - 2) + (-4s + 11)/(s^2 + 2s + 10)] = (1/9) [4/(s - 2) + (-4s + 11)/((s + 1)^2 + 9)], by completing the square = (1/9) [4/(s - 2) + (-4(s + 1) + 15)/((s + 1)^2 + 9)] = (1/9) [4/(s - 2) - 4 (s + 1)/((s + 1)^2 + 9) + 5 * 3/((s + 1)^2 + 9)]. Inverting term by term yields (1/9) [4e^(2t) - 4e^(-t) cos(3t) + 5e^(-t) sin(3t)].

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