rojsaidsadq996
rojsaidsadq996 rojsaidsadq996
  • 07-05-2017
  • Mathematics
contestada

inverse laplace of [(1/s^2)-(48/s^5)]

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LammettHash
LammettHash LammettHash
  • 08-05-2017
[tex]\mathcal L^{-1}_t\left\{\dfrac1{s^2}-\dfrac{48}{s^5}\right\}=\mathcal L^{-1}_t\left\{\dfrac{1!}{s^2}\right\}-2\mathcal L^{-1}_t\left\{\dfrac{4!}{s^5}\right\}=t-2t^4[/tex]

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