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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    factor s^3 out of the bottom and then use partial fractions answer: 2t^2+3sin2t

  2. anonymous
    • 5 years ago
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    How factor s^3 out of the bottom?

  3. anonymous
    • 5 years ago
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    How can you not, there's a common factor of s^3

  4. anonymous
    • 5 years ago
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    OKKK! 2(3s^3+2s^2+8)/(s^2+4)s^3

  5. anonymous
    • 5 years ago
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    then use partials and proceed as normal

  6. anonymous
    • 5 years ago
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    2(3s^3+2s^2+8)=A(S^2+4)+B(S^3) is correct?

  7. anonymous
    • 5 years ago
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    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

  8. anonymous
    • 5 years ago
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    (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)

  9. anonymous
    • 5 years ago
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    yeah that's right not take inverse laplace

  10. anonymous
    • 5 years ago
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    simplifying and inv I have to get your answer =2t^2+3sin2t, right????????? thankuuuuuu

  11. anonymous
    • 5 years ago
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    yeah

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