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Separable DE

  • Thread starter cscott
  • Start date
  • #1
782
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Homework Statement



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]

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
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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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.
 
  • #3
782
1
EDIT: thinking
 
Last edited:
  • #4
782
1
You missed the 'k' when solving for C'

[tex]C' = -\frac{m}{kC}e^{kv_0}[/tex]
 
  • #5
782
1
Think I got it
 
Last edited:
  • #6
782
1
Thanks
 

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