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I have to solve the differential equation: $v\,'=g-cv$.

Sorry in advance for lack of latex. I will learn it soon, please let me make a question using the common programming notation for my expressions, like * for product, ^ for exponentiation, ' for derivative.

Where $v$ is the speed, so it has distance/time units, so its derivative has acceleration units, $g$ is acceleration and $c$ is a constant with units 1/time, so the 3 terms have the same units.

I'm not worried right now about finding all the solutions, or finding maximal interval of existence, I'm worried about the units of the solution that came out of my equation, this is my approach to solution:

1)$v\,'=g-cv$ the separation of variables
2)$\displaystyle\frac{v\,'}{g-cv}=1$ then integration with respect to variable $t$ ( assume $v=v(t)$ )
3)$\displaystyle\frac{ln(g-cv)}{-c}=t+K$ with $K$ a constant with time units.

Until now, both sides match their units. I'm a little worried about taking the log of g-cv, but I keep going, solving for v:

4)$\displaystyle v=\frac{g-e^{-c(t+k)}}{c}$

Now I have an acceleration g adding to an exponential, wich seems to me is unit-less.

What is wrong with this? Can i do what I did or some steps are wrong?

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  • $\begingroup$ what is $g$? a constant of function of $t$ ? $\endgroup$ – Chiranjeev_Kumar Jul 27 '15 at 15:45
  • $\begingroup$ @Chiranjeev Yes, g is a constant with units of acceleration. It represents the acceleration gravity at earth's surface, so it's a constant, in fact it's aproximately 9.8 m/s^2. $\endgroup$ – Santropedro Jul 27 '15 at 17:32
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There are no troibles with units. Look, we find solution of $$ \frac{dv}{g-cv} = dt; $$ ok, integrate it: $$ \int_{v_0}^v \frac{dv}{g-cv} = \int_{t_0}^{t} dt\Longrightarrow \ln\frac{g-cv}{g-cv_0}=-c(t-t_0). $$ As you can see, both log and $ct$ are unitless. We can express $v$: $$ \ln\frac{g-cv}{g-cv_0}=-c(t-t_0) \Longrightarrow v=\frac gc - \left(\frac gc-v_0\right)e^{-c(t-t_0)} $$

EDIT (answer to comment). There is some trick. In math, if we have equation $$ y'=f(y)g(x), $$ we simply write $$ \frac{dy}{f(y)}=g(x)\,dx. $$ It's true regardless units. But for unitless variables we can write $$ \int\frac{dy}{f(y)}=\int g(x)\,dx + C; $$ this constant $C$ is unitless too. But it's short notation for $$ \int_{y_0}^y \frac{dy}{f(y)}=\int_{x_0}^x g(x)\,dx. $$ Roughly speaking, actually we write $$ \int\frac{dy}{f(y)} + C_1 =\int g(x)\,dx + C_2, $$ and $C_1$ and $C_2$ are unit constant. In unitless case, $C=C_2-C_1$. In your case, $K$ is not a time. I hope, this explanation will be useful.

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  • $\begingroup$ Please, solve for v in your answer, and THEN the problem, wich on my solution is on step 4, will appear. You left a implicit equation for v, solve it here and the problem will appear. $\endgroup$ – Santropedro Jul 27 '15 at 17:22
  • $\begingroup$ @user1968296, see my edit. Check your calculations $\endgroup$ – Michael Galuza Jul 27 '15 at 17:35
  • $\begingroup$ You solved it correctly, there is no doubt about that. But I don't know what's wrong in my calculations. I did it in a different way than you, I normally put in the general solution and then i plug the initial values. I used this method to solve unit-less equations probably when you solve equation with units, it's important to do it that way. I have "units" inside my logarithm, probably I can do that, that would make sense. Thank you very much. $\endgroup$ – Santropedro Jul 27 '15 at 17:44
  • $\begingroup$ @user1968296, see my edit $\endgroup$ – Michael Galuza Jul 28 '15 at 2:55

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