2
$\begingroup$

I am trying to learn more about calculus by myself, in order to be able to use dynamical systems analysis methods. In a book example, I have to find $f(t)$ from this:

$$\frac{df(t)}{dt}+\frac{v}{V}f(t) =\frac{m}{V}$$

with $m$, $v$ and $V$ being parameters

This is a linear differential equation, so I used an integrating factor (as advised in the book) and I get:

$$\frac{d\left(f(t)\cdot \exp\left[\frac{v}{V}t\right]\right)}{dt}=\frac{m}{V}\cdot \exp\left[\frac{v}{V}t\right]$$

I know what the answer is supposed to be, but I keep failing at finding the good solution. I guess my knowledge and skills are too low, but I am working on it and will edit :)

Could you please tell me what are the steps to integrate this and get an expression of $f(t)$?

According to the book the solution is : $$f(t)=\frac{m}{v}-k e^{-\frac{v}{V} t}$$

And it is also the output I get from Mathematica from this code:

DSolve[f'[t] + (v/V)*f[t] == m/V, f[t], t]

I still cannot find that on paper :/

Here is what I find (please don't hate about the maths and LaTex, I just started learning):

$$\int {d\left(f(t)\cdot \exp\left[\frac{v}{V}t\right]\right)}=\frac{m}{V}\int \cdot \exp\left[\frac{v}{V}t\right]\cdot dt$$

which gives

$$f(t)\cdot \exp\left[\frac{v}{V}t\right]=\frac{m}{V}\cdot\frac{V}{v}\cdot \exp\left[\frac{v}{V}t\right]$$

so

$$ f(t)=\frac{m}{v} $$

WHERE AM I WRONG ?

$\endgroup$
  • $\begingroup$ Integrate both sides of the last equation with respect to $t$ so you will have on the left hand side $e^{\frac{v}{V}t}f(t)$. $\endgroup$ – Mhenni Benghorbal Jul 30 '13 at 11:47
  • $\begingroup$ I did that but then I end up with f(t)=m/v which is wrong :/ $\endgroup$ – Ouistiti Jul 31 '13 at 15:03
  • 2
    $\begingroup$ Don't forget your $+C$ when integrating. $\endgroup$ – littleO Aug 1 '13 at 8:51
  • $\begingroup$ I was so blind !!! This solve everything lol. Can I get the stupid badge ? Haha. Thanks ! (Actually I was using what was in the answer below, and there was no constant) $\endgroup$ – Ouistiti Aug 1 '13 at 8:55
2
$\begingroup$

Following on from your work, we have $$e^{\frac{v}{V}t}f(t) = \int\frac{m}{V}e^{\frac{v}{V}t}dt.$$ As $\frac{m}{V}$ is a constant, it can be taken out the front of the integral, so we just need to find the antiderivative of $e^{\frac{v}{V}t}$, that is, a function with derivative $e^{\frac{v}{V}t}$. As $\frac{v}{V}$ is a constant, we use the fact that for $k \neq 0$, we have $\int e^{kt}dt = \frac{1}{k}e^{kt}$. By choosing $k$ appropriately, you should be able to obtain the antiderivative of $e^{\frac{v}{V}t}$. From there you should be able to find $f(t)$.

$\endgroup$
0
$\begingroup$

So it was ridiculously simple, just omitted the constant when integrating...

$$\int {d\left(f(t)\cdot \exp\left[\frac{v}{V}t\right]\right)}=\frac{m}{V}\int \cdot \exp\left[\frac{v}{V}t\right]\cdot dt$$

which gives

$$f(t)\cdot \exp\left[\frac{v}{V}t\right]+C=\frac{m}{V}\cdot\frac{V}{v}\cdot \exp\left[\frac{v}{V}t\right]$$

so we divide everything by $exp\left[\frac{v}{V}t\right]$

$$ f(t)+\frac{C}{exp\left[\frac{v}{V}t\right]} =\frac{m}{v} $$

$$ f(t) =\frac{m}{v}-C\cdot \exp\left[-\frac{v}{V}t\right]$$

Thank you littleO

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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