Is my explanation of why we can make $\frac{dy}{dx} {dx} = {dy}$ in the separable differential equations valid?

Question

I'm trying to understand why we can separate $\frac{dy}{dx}$ in separable differential equations when rigorously it's not a fraction but an operator (AFAIK).

I have an assumption, but I'm not sure that it's right. I'm not a mathematician, just a math-curious self-taught programmer and factory worker (all together). I eventually realized that I don't understand how the separation of $\frac{dy}{dx}$ actually works in the background and got kind of nerd-sniped (xkcd 356). I never took a course on rigorous Analysis, just lectures on Calculus a few years ago, from MIT OCW and Professor Leonard and such.

Well, I came here to the community of mathematicians to ask if my first attempt is correct and what should I do with the second one in order to finish it, because I am stuck there.

Fist attempt:
$\frac{dy}{dx} = f(x)g(y)$, $\frac{1}{g(y)}\frac{dy}{dx} = f(x)$, we can write: $\frac{1}{g(y)}$ as: $\frac{d}{dy} \int\frac{1}{g(y)}dy$.

Then: $\frac{d}{dy} \int\frac{1}{g(y)}dy\frac{dy}{dx} = f(x)$. I see a similarity to the chain rule, so maybe I can use it. Recall "The Chain Rule": $\frac{du}{dx} = \frac{du}{dy} \frac{dy}{dx}$, and let $u = \int\frac{1}{g(y)}dy$, so if we substitute it into the chain rule, we get: $\frac{d}{dx} \int\frac{1}{g(y)}dy = \frac{d}{dy} \int\frac{1}{g(y)}dy \frac{dy}{dx}$, since the first term of the right-hand side is simply a differentiation of an integration with respect to the same variable ${y}$ then according to the fundamental theorem of calculus:
$\frac{d}{dx} \int\frac{1}{g(y)}dy = \frac{1}{g(y)} \frac{dy}{dx} = f(x)$.

Now if I integrate all this with respect to $x$: $\int[\frac{d}{dx} \int\frac{1}{g(y)}dy]dx = \int[\frac{1}{g(y)} \frac{dy}{dx}]dx = \int f(x) dx$. I finally get $\frac{dy}{dx}$ separated without "abusing notation": $\int\frac{1}{g(y)}dy = \int\frac{1}{g(y)} \frac{dy}{dx}dx = \int f(x) dx$; $\int\frac{1}{g(y)}dy = \int f(x) dx$.

Is this attempt correct?

Second attempt (which I stuck with at the end):
$\frac{dy}{dx} = f(x)g(y)$; $\frac{1}{g(y)}\frac{dy}{dx} = f(x)$.

Let: $G(u) = \int\frac{1}{g(u)}du$$ where ${u} = y(x)$ and therefore ${du} = \frac{d}{dx}y(x)$. Then I think that I can rewrite: $G(u) = \int\frac{1}{g(u)}du$ as: $G(y(x)) = \int\frac{1}{g(y(x))} \frac{d}{dx} y(x)$.

Then: $\frac{1}{g(y)}\frac{dy}{dx} = f(x)$ can be rewritten as: $\frac{d}{dy}G(y(x)) \frac{dy}{dx} = f(x)$.

Recall the Chain Rule: $\frac{du}{dx} = \frac{du}{dy} \frac{dy}{dx}$ and in this case the $u = G(y(x))$, so we get: $\frac{d}{dx} G(y(x)) = \frac{d}{dy} G(y(x)) \frac{dy}{dx} = f(x)$, then I integrate all this with respect to ${x}$ like this:

$\int\frac{d}{dx} G(y(x)) dx = \int[\frac{d}{dy} G(y(x)) \frac{dy}{dx}]dx = \int f(x) dx$, since $G(y(x)) = \int\frac{1}{g(y(x))} \frac{d}{dx} y(x)$:
$\int[\frac{d}{dx} \int\frac{1}{g(y(x))} \frac{d}{dx} y(x)] dx = \int[\frac{d}{dy} \int\frac{1}{g(y(x))} \frac{d}{dx} y(x) \frac{dy}{dx}]dx = \int f(x) dx$

And here I stuck and don't know what (and why) to do with all this mess.

Answer 1

Let $$\frac{dy}{dx}=f(x)g(y)$$ be the generic separable ODE. Let's mess around unrigorously solving it (I spare us the well-known details): $$ \int_{x_0}^xf(u)\,du=\int_{y_0}^y\frac1{g(u)}\,du\,. $$ In the notation $$ F(x,y):=\int_{x_0}^xf(u)\,du-\int_{y_0}^y\frac1{g(u)}\,du=0 $$ this defines the solution $y(x)$ as an implicit function.

Theorem. This implicit function solves the original ODE.

Proof. $$ 0=\frac{d}{dx}F(x,y)=\frac{\partial}{\partial x}F(x,y)+\frac{\partial}{\partial y}F(x,y)\,\frac{dy}{dx}=f(x)-\frac 1{g(y)}\frac{dy}{dx}\,. $$ $$\tag*{$\Box$} $$ Remark. This proof just mimics what every person solving and ODE should do: verifying the obtained solution.

Answer 2

You can make a change of variables in your $G(u)$, but you are missing the $dx$ in the integral formula for $G(y(x))$.

Your concern was expressed by the well-known educator David Tall in the following terms:

Students are often given emotionally charged instructions to avoid thinking of $dy/dx$ as a quotient and to conceptualize it as a limit, even though the formulae of the calculus visibly seem to operate as if it is a quotient involving symbols that can be shifted around to change differential equations into integrals. (Tall on page 336 in Tall, David. How Humans Learn to Think Mathematically: Exploring the Three Worlds of Mathematics. Cambridge University Press, 2013)

In fact, to give a convincing account of the separation of variables technique, one can use a modern theory of infinitesimals such as nonstandard analysis. Such an account still requires nontrivial work. You could consult Keisler's book Foundations of infinitesimal calculus.

Answer 3

I know this isn't rigorous, but at its core, dy IS some infinitesimally small change in y, and dx is some infinitesimally small change in x. The derivative operator $\frac{d}{dx}$ just examines the infinitely small change in something compared to an infinitely small change in x. (As an aside: $\frac{dy}{dx}$ is NOT an operator. $\frac{d}{dx}$ is. $\frac{dy}{dx}$ is the result you get out after applying $\frac{d}{dx}$ to y)

So, when thought of that way, when you apply the operator $\frac{d}{dx}$ to y, you do get a fraction of $\frac{dy}{dx}$. The number you get out represents the ratio between dy and dx at a given point. And so dy and dx are theoretically measurable and that's the reason $\frac{dy}{dx}$ equals a number when evaluated at a point.

All this to say, that's why you can treat $\frac{dy}{dx}$ as a fraction and manipulate it as such.

As another aside, you cannot do the same thing with $\frac{d}{dx}$ because $\frac{d}{dx}$ is the operator itself that says "take the derivative of something with respect to x", or "find an infinitely small change in something and compare it to an infinitely small change in x." Multiplying out the dx wouldn't make sense because it's an unevaluated operator. At least, that's what helps me sleep at night.