Separable DE

1. May 24, 2008

cscott

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

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

3. The attempt at a solution

$$m\dot{v} = -Ce^{-kv}$$

Throwing this in Maple gives me

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

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. May 24, 2008

HallsofIvy

Staff Emeritus
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
$$v(0)= v_0= \frac{1}{k}ln\left(\frac{-kC(C')}{m}\right)$$
which gives
$$\frac{-kC(C')}{m}= e^{kv_0}$$
or
$$C'= -\frac{m}{C}e^{kv_0}$$
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. May 24, 2008

cscott

EDIT: thinking

Last edited: May 24, 2008
4. May 24, 2008

cscott

You missed the 'k' when solving for C'

$$C' = -\frac{m}{kC}e^{kv_0}$$

5. May 24, 2008

cscott

Think I got it

Last edited: May 24, 2008
6. May 24, 2008

Thanks