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I was solving a physics problem about a falling element that gets air resistance.
In that case, the acceleration, which is the derivative of velocity, is defined as $$a=\frac{mg-bv}{m}$$ And, m,g,b is all a constant.
Let's define a function $v(t)=\text{the velocity at the time t}$
Then, we can define the derivative of velocity($a$) with the expression written above.
Like this. $$\frac{dv(t)}{dt}=\frac{mg-bv(t)}{m}$$ Now, we can define the physics problem I was solving by pure math.

Abstracted problem
function v(t) is defined as $$v(t)=\frac{mg-m\frac{dv(t)}{dt}}{b}$$ Can v(t) be defined without using it's derivative? If it can't, why? If it can, how can it be done, and what is the definition?

I am new pretty new to calculus and also physics. This question might be a stupid or a simple question. In that case, sorry. Thanks in advance.

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    $\begingroup$ Could you explain why you swapped the positions of $v(t)$ and $\frac{\mathrm{d}v(t)}{\mathrm{d}t}$ between the second and third display equations? $\endgroup$ – Eric Towers Dec 7 '19 at 7:22
  • $\begingroup$ @EricTowers I have no special reason of that. Just wanted to define the function by the derivative. $\endgroup$ – dissolve Dec 7 '19 at 7:27
  • $\begingroup$ @EricTowers It appears OP solved for $v(t)$ in the third display (note the denominator also changed). $\endgroup$ – Teepeemm Dec 7 '19 at 15:50
  • $\begingroup$ @Teepeemm : Ah. Thanks! $\endgroup$ – Eric Towers Dec 7 '19 at 15:52
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We have that

$$a=\frac{mg-bv}{m} \iff a=g-\frac b mv$$

and by $u=g-\frac b mv$ $$\frac m b \dot u=\frac m b (0-\frac b m u)=-u \implies \frac m b \frac{du}{dt}=-u \implies \frac m b \frac{du}{u}=-dt$$$$\implies \frac m b \ln u=-t +C' \implies u=e^{-\frac b mt+C'}=Ce^{-\frac b mt}$$

that is

$$g-\frac b mv=Ce^{-\frac b mt}$$

$$v=\frac m b g-\frac m bCe^{-\frac b mt}$$

and by $v(0)=v_0$ we obtain $C=g-\frac b m v_0$.

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    $\begingroup$ Doesn't it has to be $v=\frac m b g-\frac m bCe^{\frac{-b}{m}t}$...? $\endgroup$ – dissolve Dec 7 '19 at 7:50
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    $\begingroup$ @dissolve Yes of course! I forgot a minus sign in the first step! I fix that. $\endgroup$ – user Dec 7 '19 at 7:52
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    $\begingroup$ @dissolve The physical interpretation helps a lot in this case! Good catch! Bye $\endgroup$ – user Dec 7 '19 at 7:53
  • $\begingroup$ Also, I cannot understand the reason why $\frac m b \dot u=-u$. Could you explain it? $\endgroup$ – dissolve Dec 7 '19 at 7:57
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    $\begingroup$ @dissolve We have defined $$u=g-\frac b mv \implies u'=0-\frac b m v'=-\frac b m a \implies a=- \frac m b u'$$ and the equation becomes $$\frac m b u'=-u$$ $\endgroup$ – user Dec 7 '19 at 8:03
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Yes, it can solved. What you have is a differential equation and the standard techniques for solving them give you that$$v(t)=\frac{gm}b-Ke^{-bt/m},$$for some constant $K$. Of course, $K=\frac{gm}b-v(0)$.

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