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Differential equation

  1. Apr 18, 2014 #1

    kyu

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    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution

    i got k = ln |9/64|

    then how can the next step using ln 0 doesn't make sense. what should i do?
     

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  3. Apr 18, 2014 #2

    Mark44

    Staff: Mentor

    Please show us your work.
     
  4. Apr 18, 2014 #3

    kyu

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    should be wrong but here goes

    dh/dt = -k h^(1/2)
    1/h^(1/2) = -k dt
    ln 9 - ln 64 = k
    ln |9/64| = k

    ln 0 - ln 64 = ln |9/64| (t-0)
     
  5. Apr 18, 2014 #4

    Mark44

    Staff: Mentor

    Where did the dh go?
    How did you get this (above)?
     
  6. Apr 18, 2014 #5

    pasmith

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    Homework Helper

    The integral of [itex]h^{-1/2}[/itex] is given by the usual rule for powers, [itex]\int x^\alpha\,dx = \frac{x^{\alpha + 1}}{\alpha + 1} + C[/itex], not the exception [itex]\int x^{-1}\,dx = \ln |x| + C[/itex].
     
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