I don't even know if this is the right place to post this, but everything I've found online so far hasn't helped. I'm currently a freshman in college taking a semi-introductory physics course (my AP physics credits got me out of the actual introduction). Today we quickly went over drag, where my professor went over solving for $v(t)$ from the differential equation $m\frac{dv}{dt} = mg - bv$, where b is a constant. I took detailed notes on what he did:
$$m\frac{dv}{dt} = mg - bv$$ $$m\frac{dv}{dt} = -b(\frac{-mg}{b} + v)$$ $$\text{Let} \; (\frac{-mg}{b} + v) = \tilde{v}, \; \text{where} \; \frac{dv}{dt} = \frac{d\tilde{v}}{dt}$$ $$m\frac{d\tilde{v}}{dt} = -b\tilde{v}$$ $$\frac{d\tilde{v}}{dt} = \frac{-b}{m}\tilde{v}$$ $$\tilde{v}(t) = Ce^{\frac{-b}{m}t}$$ $$v(t) = Ce^{\frac{-b}{m}t} + \frac{mg}{b}$$
I get 100% lost when he substitutes $\tilde{v}$ for $\frac{-mg}{b} + v$. Why can we assume the derivative of $v$ is the same as the derivative of $\tilde{v}$, and how does integrating with respect to $t$ give us $Ce^{\frac{-b}{m}t}$? If anyone could explain this to me I would very much appreciate it. I have an exam this weekend and this could be on it, so I'm kind of stressed about it.
The derivative of $\tilde{v}$ and that of $v$ are the same because $\tilde{v}$ and $v$ differ only by $mg/b$ which doesn't depend on $t$.
Getting the $y=Ce^{kt}$ out of $\frac{dy}{dt}=ky$ can be done in a number of different ways. One of them is separation of variables, which can be written like this:
$$\frac{1}{y} \frac{dy}{dt} = k \\ \int \frac{1}{y} \frac{dy}{dt} dt = \int k dt \\ \ln(|y|)=kt+C \\ |y|=e^C e^{kt} \\ y=\pm e^C e^{kt}.$$
Then identify $\pm e^C$ as an arbitrary nonzero constant say $C'$. Finally, check that $y=0$ is also a solution, so that $y=C e^{kt}$ is a solution for any real number $C$. (Usually this is presented in a less technically correct way in textbooks.)
Incidentally, problems along these lines are one of the main topics in "elementary differential equations", which in the "standard" US undergraduate math sequence comes immediately after calculus III.