Treating dy/dx as a symbol for the derivative as opposed to a fraction of the differentials dy and dx

Question

I have some questions in solving the seperable differential equation $dy/dx=x/y$. But first I'll explain what I understand about the problem and how it relates to the differentials $dy$ and $dx$. Correct me if I am wrong.

I understand that we define a dependent variable $dy=dy/dx*dx$ where $dy/dx$ is the single symbol representing the derivative and $dx$ is an independent variable that can be chosen freely. By solving this equation for $dy/dx$ (the derivative) we get $dy/dx$ (the fraction). This allows us to freely choose any expression for $dx$ and by definition we will get the exact expression for $dy$ so that $dy/dx$ (the fraction) is the same as the single symbol $dy/dx$ (the derivative).

In textbooks we seem to skip the justification above and immediately interpret the single symbol $dy/dx$ as fraction in certain situation.

Questions...

1) Why don't we at least say... "hey... by our study on differentials in an earlier section... we will interpret $dy/dx$ as a fraction in this next step"? This was always so confusing to me in school when we applied it without saying anything. I didn't understand why there was cancellation. I do now. In any case, it does makes sense to me now that the variables cancel and we're left with $y*dy = x*dx$ where $dx$ and $dy$ are variables.

2) So now we are ready to integrate both sides. But what are we integrating with respect to if both $dy$ and $dx$ are just variables? It seems that we are treating them differently again, but what's the justification? If if there is some why don't we state it anywhere?

3) Wouldn't it make more sense to solve the problem this way. Multiply both of $dy/dx=x/y$ by y only. This will yield the equation $y*dy/dx=x$. Then we can integrate both side with respect to x, applying the chain rule on the left side?

I know we get the same result by separating the $dx$ and the $dy$ on different sides...and then treating $dx$ and $dy$ as the the part of the integral that tells us which variable we are integrating with respect to... But this seems fuzzy to me as they are really variables.

Isn't (3) more technically correct?

Answer 1

Definitely, $(3)$ is more rigorously correct.
1) Yes, in many books they vaguely use the symbols $dy$ and $dx$ as if they are normal variables. As in your example $$\frac{dy}{dx}=\frac{x}{y}$$ it is just to say that the derivative of the function $y$ with respect to the independent variable $x$ is a function of $x$ and $y$. Nothing more to it, and yes, it is incorrect to manipulate $\frac{dy}{dx}$ as a fraction. You can visit this link to really see that manipulating $dy$ and $dx$ as fractions is just a matter of notation.

2)We are integrating with respect to $x$, that is the independent variable. $y$ is a function of $x$, and $\frac{dy}{dx}$ is its derivative with respect to $x$. As you can see in your example: $$\begin{align}\frac{dy}{dx}=\frac{x}{y} \\ \implies y\frac{dy}{dx}=x \end{align}$$ We have the derivative of the function $y$ along with some expression of $y$.So, chain rule can be used to integrate the LHS w.r.t $x$, and for the right, it is just a function of $x$, so integrating it w.r.t $x$ is straightforward. $$\int y\frac{dy}{dx}dx=\int xdx$$ Here is the abuse of notation we use. As you can see, to integrate the LHS, we integrate normally as if $y$ were the independent variable, as the chain rule suggests. So, the notation $$\frac{dy}{dx}dx$$ is just hacked to cancel $dx$, to make the name "Variable separable form". That's it, its just notation.

Also, even though I don't know much about the modern theory of infinitesimals, I think they do provide us mathematically justified steps to show why the symbols $dy$ and $dx$ work as normal fractions.