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anonymous
 5 years ago
Please helpme wid inverse Laplace of
(6s^3+4s^2+16)/(s^5+4s^3)
thanx
anonymous
 5 years ago
Please helpme wid inverse Laplace of (6s^3+4s^2+16)/(s^5+4s^3) thanx

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0factor s^3 out of the bottom and then use partial fractions answer: 2t^2+3sin2t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How factor s^3 out of the bottom?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How can you not, there's a common factor of s^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OKKK! 2(3s^3+2s^2+8)/(s^2+4)s^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then use partials and proceed as normal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02(3s^3+2s^2+8)=A(S^2+4)+B(S^3) is correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.0(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
 5 years ago
Best ResponseYou've already chosen the best response.0yeah that's right not take inverse laplace

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0simplifying and inv I have to get your answer =2t^2+3sin2t, right????????? thankuuuuuu
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