function defined by its derivative at a specific point

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

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