Understanding a simple differential equation using "separation of variables" On pg. 44 of No-Nonsense Classical Mechanics, the author gives an example of solving a differential equation via "separation of variables":




*

*In the fourth step (where the $\int$ is added), what is the literal meaning of $dx(t)$?  In the integral, does $dx(t)$ literally expand to $x'(t)dt$?


*In the fifth step, why is $\int C dt = Ct + c$? Isn't it just $\int C dt = Ct$ (i.e., where does the $+ c$ come from)?


*Also in the fifth step, why does $\int \frac{dx(t)}{x(t)} = \ln(x(t))$?
 A: ${(1)}$ Well, technically we shouldn't treat ${dx(t)}$ or ${dt}$ as numbers. But in this context, yes, we are treating ${dx(t)}$ as ${x'(t)dt}$. It's just a rearrangement of ${\frac{dx(t)}{dt} = x'(t)}$.
${(2)}$ The ${+c}$ is required. When you integrate indefinitely, you always need to add a constant. This is because ${\frac{d}{dx}(f(x)+c)=\frac{d}{dx}f(x)}$ for any constant ${c \in \mathbb{R}}$, and the integral is meant to be a sort of "inverse" to the derivative. If I tell you I thought of a function, and the derivative of that function is ${2x}$ - did I think of ${x^2+1}$ or just ${x^2}$? You don't know, and there is no way to know without further information. Hence you must say "you thought of ${x^2 + c}$ for some constant $c$, but I don't know what the constant you chose is".
You may ask: "why isn't there a ${+c}$ on the left hand integral then too?" and that's because the constant on the left could just be moved to the other side and you would end up with "constant - constant", which is just another constant. So yes, we did have to add a "${+c}$" to both sides, but the ${+c}$ on one side can just be absorbed into the constant on the other side.
${(3)}$ Well, as you said, ${dx(t)=x'(t)dt}$. So the integral becomes
$${\int \frac{x'(t)dt}{x(t)}}$$
And after substituting ${u=x(t)}$ you get
$${\int \frac{du}{u}=\ln(u)+c}$$
But since ${u=x(t)}$ then
$${=\ln(x(t))+c}$$
Edit: I actually saw your question posted here: What is the meaning of the differential of a variable when the variable is the value of a constant function? with your example of ${\int_{1}^{2}df(t)}$. What this evaluates to actually depends on what you mean by your bounds. In other words, do you mean
$${\int_{t=1}^{t=2}df(t)}$$
or do you mean
$${\int_{f=1}^{f=2}df(t)}$$
? If you mean the first, the answer is
$${\int_{t=1}^{t=2}f'(t)dt=f(2)-f(1)}$$
And so obviously the fact $f$ is a function of $t$ really does matter. If you mean the second, then
$${\int_{f=1}^{f=2}df(t)=\int_{f=1}^{f=2}df=2-1=1}$$
This may confuse you as (int the pre-edit version of my answer) I said it didn't matter - and that ${\int \frac{dx(t)}{dx}=\int \frac{dx}{x}}$ will give you the same answer. The reason it didn't matter is because we hadn't picked any actual bounds yet, and things get kinda "hidden away" by ${u}$ substitution. The proof of separation of variables actually proves for us that ${\frac{dy}{dx}=\frac{f(x)}{g(y)}}$ implies that
$${\int g(y)dy = \int f(x)dx}$$
by using a $u$ substitution. So technically, we cannot just say ${\int \frac{dx(t)}{x(t)}=\int \frac{dx}{x}}$ - it requires further proof, and the proof relies on substitution. Meaning the first way I mentioned is a bit circular, and I have removed it from my answer. Substitution is the proper way - but you don't need to do this everytime.
