anonymous
  • anonymous
Please helpme wid inverse Laplace of (6s^3+4s^2+16)/(s^5+4s^3) thanx
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
factor s^3 out of the bottom and then use partial fractions answer: 2t^2+3sin2t
anonymous
  • anonymous
How factor s^3 out of the bottom?
anonymous
  • anonymous
How can you not, there's a common factor of s^3

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anonymous
  • anonymous
OKKK! 2(3s^3+2s^2+8)/(s^2+4)s^3
anonymous
  • anonymous
then use partials and proceed as normal
anonymous
  • anonymous
2(3s^3+2s^2+8)=A(S^2+4)+B(S^3) is correct?
anonymous
  • anonymous
I'm assuming that's your set up of the partial and you haven't solved the partial fraction yet. If so, that set up is incorrect because s^2+4 is a irreducible quadratic and s^3 is repeated 3 times.. Though, when you actually solve the partial fraction you will find the other constants are actually 0 so it is indeed left in the form you put it as, but I believe you do need to solve for those 0 constants in order to find your remaining 2 missing coefficients anyways
anonymous
  • anonymous
(6s^3+4s^2+16)=A(S^2+4)+B(S^3); A=4, B=6 (6s^3+4s^2+16)/(s^5+4s^3)= 4/S^3 + 6/(S^2+4)
anonymous
  • anonymous
yeah that's right not take inverse laplace
anonymous
  • anonymous
simplifying and inv I have to get your answer =2t^2+3sin2t, right????????? thankuuuuuu
anonymous
  • anonymous
yeah

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