efigu5edjimAnis efigu5edjimAnis
  • 21-11-2016
  • Mathematics
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(2s+5)/(s^2+6s+34) inverse Laplace transform

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syed514
syed514 syed514
  • 23-11-2016
(2s+5)/(s^2+6s+9+25) = (2s+6-1)/((s+3)^2+25) = (2(s+3)-1)/((s+3)^2+5^2) = 2[(s+3)/((s+3)^2+5^2) - 1/5 (5 / ((s+3)^2+5^2))
Hence inverse Laplace transform is 2e^(-3t)cos 5t - (1/5) e^(-3t)sin 5t
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