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Differential equation of growth & decay

  1. Feb 22, 2012 #1
    Question: find the rate of change of (s) with respect to time(t), is inversely proportional to the square root of (s)

    Write a differential equation for this statement.

    Find the general solution to this equation

    If initially (s)= 100, and after six seconds (s)= 144, what is the value of (s) be after 10th seconds?

    Work so far:
    Part one, ds=k/sqrt(s) dt

    Part two, sqrt(s) ds = k dt
    2/3(s)^3/2 = kt+c
    (s)^3/2 = 3/2 kt +c
    S=(3/2 kt +c)^2/3
    So far the above is correct, and I know that these are true
    T=0, s=100
    T=6, s=144
    T=10, s=?
    The final answer is s=(6640/3)^2/3
    I just don't know what to do to get the final answer.....
     
  2. jcsd
  3. Feb 22, 2012 #2

    vela

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    Use the information given to solve for k and c. For example, you're given s(0)=100. You also know s(0) = [3/2 k(0) + c]2/3. Putting those two together, you can solve for c.
     
  4. Feb 22, 2012 #3
    100^3/2 = c
    Ergo 1000= c
    144=(3/2 k 6 + 1000)^2/3
    144^3/2 -1000= 9k
    (144^3/2 -1000)/9=k
    S=(3/18(144^3/2-1000)*10 +1000)^2/33
    And I need a calculator to check that. Last time I did it the way, my answer did not match the final answer...
     
  5. Feb 22, 2012 #4

    vela

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    ##144^{3/2} = (\sqrt{144})^3 = 12^3 = 1728##
     
  6. Feb 22, 2012 #5
    Thanks!! Am a bit tired at the moment.
    1728-1000= 728728
    728*10= 7280
    3/18*7280= 3640/3 + 3000/3
    (6640/3)^3/2 which is the correct answer :-D
    Thanks!!
     
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