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Calculus help

  1. Jun 9, 2005 #1
    y''= -3/(t^3)

    s=0 when t = 1
    s'= -3/2 when t= (3)^1/2

    my answer is....

    -3/(2t^-2) - 2t + 7/2

    please tell me if that is correct
  2. jcsd
  3. Jun 9, 2005 #2


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    Not really.You should have integrated twice correctly.

    [tex]y'=-3\frac{t^{-2}}{-2}+... [/tex]

  4. Jun 9, 2005 #3


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    First, you have "y" in the differential equation but "s" in the initial conditions. Which is it?
    Second, you have -3/(2t^-2) in your answer which would be the same as (-3/2)t^2
    Do you mean (-3/2)t^(-2)= -3/(2t^2)?

    What is the integral of -3/(t^3)= -3t^(-3)? What is the integral of that?
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