function defined by its derivative at a specific point

Question

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.

Answer 1

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 $C=g-\frac b m v_0$.

Answer 2

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, $K=\frac{gm}b-v(0)$.