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Separable DE

  1. May 24, 2008 #1
    1. The problem statement, all variables and given/known data

    boat turns off engine when it's attained [itex]v_0[/itex] at [itex]t = 0[/itex]. Starting from that moment the boat is slowed by retarding force [itex]F = Ce^{-kv}[/itex]

    3. The attempt at a solution

    [tex]m\dot{v} = -Ce^{-kv}[/tex]

    Throwing this in Maple gives me

    [tex]v(t) = \frac{1}{k}\ln\left(\frac{-kC(t + C^{'})}{m}\right)[/tex]

    This doesn't make sense to me--you can't take the log of a negative number and even if I remove that negative from the initial DE v(t)->inf as t->inf
     
  2. jcsd
  3. May 24, 2008 #2

    HallsofIvy

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    What makes you think this involves the logarithm of a negative number? Whether -kC(t+ C') is negative or not depends on C'. In particular, you know that when t= 0, v= v0. Putting this into your formula
    [tex]v(0)= v_0= \frac{1}{k}ln\left(\frac{-kC(C')}{m}\right)[/tex]
    which gives
    [tex]\frac{-kC(C')}{m}= e^{kv_0}[/tex]
    or
    [tex]C'= -\frac{m}{C}e^{kv_0}[/tex]
    which is negative. You will be taking the logarithm of a positive number as long as t< -C'. But when t= -C', what happens?


    That's the problem with using Maple, or a computer or any such rather than doing it yourself! If you have done the rather simple integral yourself you wouldn't have had that question.
     
  4. May 24, 2008 #3
    EDIT: thinking
     
    Last edited: May 24, 2008
  5. May 24, 2008 #4
    You missed the 'k' when solving for C'

    [tex]C' = -\frac{m}{kC}e^{kv_0}[/tex]
     
  6. May 24, 2008 #5
    Think I got it
     
    Last edited: May 24, 2008
  7. May 24, 2008 #6
    Thanks
     
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