Separation of variables -- what to do about the case where we might be dividing by zero? In not a few textbooks, we have some version of this example of a falling body:

Problem. Solve $\frac{dv}{dt} = a - bv$ with the initial condition $(t,v)=(0,0)$.
Solution. Rearrange to get $\frac{dt}{dv} \overset 1= \frac{1}{a - bv}.$
Integrating, $t=-\frac{1}{b}\ln|a-bv|+C_1$.
Rearranging, $e^{b(C_1-t)}=|a-bv|$ or $a-bv=\pm e^{b(C_1-t)}=C_2 e^{-bt}$.
Plug in the initial condition to get $C_2=a$ and $a-bv=ae^{-bt}$.
Now rearrange to get the solution: $v=\frac{a}{b}(1-e^{-bt}).$

My question is: At $\overset 1=$, we assume $a-bv\neq0$---so, how do we handle the case where $a-bv=0$?
(Do we for example simply assert that this never happens? How do we justify this assertion? Or is there some other more careful/precise way of dealing with $a-bv=0$?)
 A: You are correct to be suspicious of this situation. Usually, you will find a mention at the start that these calculations are only for those parts of a solution where the term you divide by is not zero. 
The zero case has to be treated separately. Fortunately in the case of autonomous first order ODE $y'=f(y)$ (with a differentiable $f$) this is easy, if $f(y_*)=0$, then $y(t)=y_*$ is a (constant) solution and by the uniqueness claim of the existence-and-uniqueness theorem no other solution can have the value $y_*$ anywhere.
Chances are that in a general solution formula that you get from the separation method or other in the non-zero case, the constant solutions are already included, for "forbidden" or "impossible" parameters of that computation. The "impossible" parameter means situations where a constant solution appears as a limit case as the parameter of the solution family goes to $\infty$. One easy example for such a situation is
$$
y'=1-y^2
\\
\frac{y'}{1+y}+\frac{y'}{1-y}=2,
\\
\left|\frac{1+y}{1-y}\right|=e^{2x+c},
\\
y(x)=\frac{Ce^{2x}-1}{Ce^{2x}+1},~~C=\pm e^c\ne 0,
$$
where you get the constant solution $y=-1$ for the "forbidden" parameter $C=0$ and the other constant solution $y=1$ for the "impossible" parameter $C=\infty$.
A: Let me first point to another issue in this method. You get to the point $|a - bv| = e^{b(C_1 - t)}$ and then introduce a new arbitrary constant $C_2$, saying that $a - bv = C_2e^{-bt}$. Think about this definition for a moment. Given that $C_2$ is supposed to be an arbitrary constant, we are tacitly permitting it to take the value $0$. What value of $C_1$ will produce $C_2 = 0$?
It's not possible; we are trying to make $C_2 = \pm e^{bC_1}$, but the right hand side can never be $0$, no matter the value of $\pm$ and $C_1$. Does this mean $C_2 = 0$ is a new, erroneous solution?
As it turns out, it is new, but it's not erroneous. It is, in fact, the solution we implicitly discarded when we divided by $a - bv$ in the first step. By dividing by this quantity, we were assuming that the function $a - bv$ was not constantly $0$ near the initial value, discarding a possible solution. But, allowing $C_2 = 0$ gives us the solution back.
I think it's always good to be careful when dividing unknown quantities, and wonder if they are $0$. In the case of separation of variables, though, it typically works out in the end, but I think it's a good thing to keep an eye on.
A: It's a good thought and one must be careful while 
dividing an expression and in general it is a good practice to 
treat these conditions as a separate case , obtain an answer and verify if the solution obtained satisfies the differential equation.
Here, if a - bv = 0 
Implies v = a/b which is constant .
Putting it in orignal equation gives LHS = RHS.
so we can safely eliminate this case for our point of study.
A: You are right to consider $a-bv=0$ as another branch of solution. No body says that the general solution must be obtained presuming $a-bv\ne 0$, unless if initial conditions force us to such. However, once $a-bv=0$, always $a-bv=0$ ! This means that if you fall on the line $v={a\over b}$, you will always remain there because you feel no force or will to leave (${dv\over dt}=0$). Also, if you are elsewhere other than $v={a\over b}$, you will never reach it, as the solution $v={a\over b}(1-e^{-bt})$ suggests. This is because the closer you get to $v={a\over b}$ asymptotically, the slower you move. This reminds a famous fact in physics of general relativity that:

A mass-less particle moving at the speed of light, cannot slow down to any speed below that of light, and a massive particle moving at some speed less that that of light, cannot move at the speed of light!

