Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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

share|improve this answer
    
I get $k(1-e^{-t})-Pv$ –  alok Aug 28 '12 at 11:31
    
So the equation is $$\frac{dv}{dt}+ckv = k(1-e^{-t}).$$ It is a linear equation as you wanted. But maybe I do not understand your question. The integrating factor is always the same: en.wikipedia.org/wiki/… –  Siminore Aug 28 '12 at 11:33
    
Yes.It's a linear equation as i wanted.Now what is $\int Pdx$> –  alok Aug 28 '12 at 11:38
    
In my opinion, you should now try to do your homework by yourself. You should compute the primitive of a constant function... –  Siminore Aug 28 '12 at 11:40
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.