Asked by mo
compute the inverse laplace transform 2S^2+13S+5/(s+3)(s-1)^2
i got 3e^t+5e^t+3e^t, not sure it correct
i got 3e^t+5e^t+3e^t, not sure it correct
Answers
Answered by
Steve
Hmm. I get
-3te^-t + 6e^-t - 4e^-3t
since (2s^2+13s+5) / ((s+3)(s+1)^2) = -3/(s+1)^2 + 6/(s+1) - 4/(s+3)
-3te^-t + 6e^-t - 4e^-3t
since (2s^2+13s+5) / ((s+3)(s+1)^2) = -3/(s+1)^2 + 6/(s+1) - 4/(s+3)
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