Find the inverse Laplace Transform of F(s)= esp(-s) / [ s(s+1)] ? Solution i am

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i am doing a sample work not meet exact criteria.. please go throuth below one Use partial fractions: 2/(s^3(s-1)^3) = A/s + B/s^2 + C/s^3 + D/(s-1) + E/(s-1)^2 + F/(s-1)^3 Clearing denominators, 2 = A(s-1)^3 s^2 + B(s-1)^3 s + C(s-1)^3 + Ds^3(s-1)^2 + Es^3(s-1) + Fs^3 s = 0 ==> 2 = C(-1)^3 ==> C = -2 s = 1 ==> 2 = F * 1^3 ==> F = 2. So, we get -6s^2 + 6s = A(s-1)^3 s^2 + B(s-1)^3 s + Ds^3(s-1)^2 + Es^3(s-1) ==> -6 = A(s-1)^2 s + B(s-1)^2 + Ds^2(s-1) + Es^2 s = 0 ==> -6 = B(-1)^2 ==> B = -6 s = 1 ==> -6 = E * 1^2 ==> E = -6. So, we get -6s^2 + 6s + 6s^3(s-1) + 6(s-1)^3 s = A(s-1)^3 s^2 + Ds^3(s-1)^2 ==> 12 = A(s-1) + Ds s = 0 ==> A = -12 s = 1 ==> D = 12. So, 2/(s^3(s-1)^3) = -12/s - 6/s^2 - 2/s^3 + 12/(s-1) - 6/(s-1)^2 + 2/(s-1)^3. Inverting term by term yields -12 - 6t - t^2 + e^t (12 - 6t + t^2). I hope this helps!

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