Suppose we have a separable equation $(t-a)x' = x$, where $x=x(t)$ and $a\in\mathbb R$ is fixed.
Per this discussion the equation has constant solutions or solutions we find via separating. Clearly, $x(t)=0, t\in D$ (more on that in a minute) is a solution. After separating we get $$\frac{1}{x}dx = \frac{1}{t-a}dt \implies x = C(t-a),\quad C\neq 0 $$ My question is about the domain of the function $x$. A solution found by separating doesn't show any problem with $t=a$, but when we separated the variables, we clearly had to assume $t\neq a$, otherwise the operation would have been illegitimate.
We have a family of straight lines as solutions. Do they now have removable discontinuities at $t=a$?. What's the story with $t=a$ or is it relevant? Can we ignore the free variable when separating?
Ok, maybe this was a lucky shot. Consider, instead $2xx'\sin t = 1$. Constant solutions are impossible and we also must have $\sin t\neq 0$. After separating we have $$2xdx = \frac{1}{\sin t}dt \implies x^2 = \ln \left\lvert \tan\frac{t}{2} \right\rvert +C$$ If we consider the first equation as $$ \frac{t-a}{x}x' = 1 $$ then comparing the two equations, in one case $t=a$ is irrelevant although initially it must be that $t\neq a$, otherwise the identity doesn't hold. In the other case $t=k\pi$ cannot hold from the get go and is a calamity for a solution.
Is it then safe to say we may ignore the free variable when separating? Is the domain of solution to be decided individually? Perhaps, there is a theorem that describes solutions to a specific equation $f(x,t,x') = c$ and there's also something assumed about the domain of $f$. I don't know what I'm looking for, though.
No, you do not have to remove discontinuities at t=a. The family of curves $$ x=C(t-a)$$ satisfies you equation $$ (t-a)x'=x$$ for every $C$ at every $t$ and that is what you are looking for.
Note that the family $$ x=C(t-a)$$ covers all the cases including the constant solution $x=0$. What I am trying to say is that once you have a solution, which satisfies your equation, you do not worry about the method which was used to get that solution.