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The equation of motion of a train is given by $m\frac{\mathrm{dv} }{\mathrm{d} t} = mk(1-e^{-t})-mcv$ where $v$ is the speed,$t$ is the time and $m,k,c$ are constants.How to find $v$ when $v=0$ and $t=0$. This is what i've tried so far $\frac{\mathrm{dv} }{\mathrm{d} t}=k(1-e^{-t}-cv)$ after seperating and factoring $m$.Now I Don't know how to put this in $\frac{\mathrm{dy} }{\mathrm{d} x}+Py = Q$ form and figure out the integrating factor.Please Help.

Thank You.

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Hint: $ k(1-e^{-t}-cv) = k(1-e^{-t}) - ckv. $ What happens if you put $P=ck$?

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    In my opinion, you should now try to do your homework by yourself. You should compute the primitive of a constant function...2012-08-28
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Let's start by re-arranging the equation:

$\dot{v} + cv = k(1-e^{-t}) \, .$

Clearly $P(t) = c$ and so the integrating factor is $\mu = e^{\int \! c \, dt} = e^{ct},$ and we shall consider

$e^{ct}\dot{v} + ce^{ct}v = ke^{ct}(1-e^{-t}) \, , $

$\frac{d}{dt} \, (e^{ct}v) = ke^{ct}(1-e^{-t}) \, , $

$e^{ct}v = k\left(\frac{e^{ct}}{c} - \frac{e^{t(c-1)}}{c-1} \right) + \alpha \, , $

$v(t) = k\left(\frac{1}{c} - \frac{e^{-t}}{c-1} \right) + \alpha e^{-ct} \, .$

Notice that $\alpha$ is a fixed constant: the constant of integration.