# How to solve this "simple" ordinary differential equation?

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 ?

• Integrate both sides of the last equation with respect to $t$ so you will have on the left hand side $e^{\frac{v}{V}t}f(t)$. Jul 30, 2013 at 11:47
• I did that but then I end up with f(t)=m/v which is wrong :/ Jul 31, 2013 at 15:03
• Don't forget your $+C$ when integrating. Aug 1, 2013 at 8:51
• 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) Aug 1, 2013 at 8:55

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 $f(t)$.

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