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Previously I asked how to integrate:

$$\frac{dy}{dx}=\frac{x}{2y}$$

Posters said that I could simply move the $2y$ and $dx$ over, essentially treating the $dx$ as a regular variable.

I wonder then why we cannot always treat differential operators as variables? I was always taught not to think of them as variables. Why not?

Michael Hardy
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Haim
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2 Answers2

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Strictly speaking, no, you cannot "move the $dx$ over". $\frac{dy}{dx}$ is not a fraction, it is a function, a full symbol in its own right, and the numerator and denominator are as inseparable as the l and the n in $\ln(x)$. However, you may move the $2y$ over, and then integrate both sides with respect to $x$ (if the function on the left-hand side is equal to the function on the right-hand side, then their integrals must also agree, up to a constant term). This gives $$ \int 2y\frac{dy}{dx} \,dx = \int x\,dx $$ Then you might recognize that what you have on the left-hand side is the chain rule for integration, which says that $$ \int 2y \frac{dy}{dx}\,dx = \int 2y\,dy $$ This allows you to solve the differential equation simply by calculating the integrals.

That being said, note that you would've gotten the same answer had you treated $dx$ as a variable and moved it to the other side. This is one of the reasons why writing derivatives as $\frac{dy}{dx}$ is so popular: even though it isn't a fraction, and the "numerator" and "denominator" aren't really terms by themselves, you still get the correct answer if you pretend that they are.

Arthur
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    The problem I've always had with people saying you can't treat $dx$ and $dy$ as separate entities is that I see them violate their own decree and write nonsense like $dy = (\partial y/\partial x_1)dx_1 + (\partial y/\partial x_2)dx_2$... excuse me, if you said you can't treat them separately then what the heck is this supposed to mean? So can you treat them as separate variables or no? Why don't mathematicians follow their own advice? I think this really deserves a good explanation somewhere... – user541686 Mar 20 '16 at 21:40
  • @Mehrdad Thing is, there are separate symbols that are called $dx$ and $dy$, but those are not the ones appearing in the fraction $\frac{dy}{dx}$. For instance, if $y = x^2$, then $\frac{dy}{dx} = 2x$, and $dy = 2x,dx$, but those two mean different things. One is a differentiation, and the other is used for integration substitution. – Arthur Mar 20 '16 at 21:48
  • But is their meaning actually different? Or is merely the situations in which they are used different? If you're really saying their meanings are different then I'd love to hear what you see as the difference. (i.e. what would be each of their definitions? This should be irrespective of their use case.) – user541686 Mar 20 '16 at 21:51
  • @Mehrdad If we take $y = x^2$, then $\frac{dy}{dx} = 2x$ is short-hand for $\lim_{h \to 0}\frac{y(x+h) - y(x)}{h} = 2x$, while $dy = 2x,dx$ is short-hand for $$\forall f: \int f(y),dy = \int 2xf(y(x)),dx,\left(= \int 2xf(x^2),dx\right)$$and connecting the two is a separate theorem called "chain rule for integrals". Only after you've proven that can you say that their meaning is similar, and exploit the fraction-like appearance of $\frac{dy}{dx}$. – Arthur Mar 20 '16 at 21:54
  • If it's just a shorthand (which I take to be an informal thing etc.) then why is it then that when I look at Wikipedia it clearly very formally says that $dy$ is a thing in itself, called the "total differential"? Nowhere does it suggest that it's just some kind of sloppy notation or shorthand... it seems to be a very formally well-defined thing in itself that I should be able to treat separately from $dx$. – user541686 Mar 20 '16 at 21:58
  • @Mehrdad Integral calculus itself is full of different interpretations and some times conflicting notions. Just ask a differential geometer or physicist how many distinct (but almost equal) things "$dx$" can mean. – Arthur Mar 20 '16 at 22:00
  • @Mehrdad Also, short-hand is in no way informal or sloppy. It is necessary to make mathematics readable. To start at the beginning, the symbol $5$ is by definition short-hand for $1+1+1+1+1$ (or in words, "The number that succeeds the number that succeeds the number that succeeds the number that succeeds the number that succeeds $0$"), and it only gets worse from there. – Arthur Mar 20 '16 at 22:04
  • I mean if you take what I said too literally (re: sloppiness etc.) then you kind of miss my actual point. I'm sure you see what I mean regarding the confusion, right? And why the explanations aren't satisfactory? Try explaining all these to a high-schooler learning calculus. You can't just tell them it's a shorthand and $dy$ has a dozen different meanings in differential geometry and then expect them to be satisfied with that answer. They won't even know what differential geometry is, but they really do have the capacity to know (and deserve to know) exactly what is going on here. – user541686 Mar 20 '16 at 22:12
  • @Mehrdad I can definitely see where your confusion comes from. Is just that all these concepts relating to "a really small change in $x$", no matter which context it's put in, gets named "$dx$", and that is definitely not fair to people just learning about this. They are the ones, more than anyone, who should be made conscious of the distinctions, and the notation is not doing then any favours. – Arthur Mar 20 '16 at 23:16
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The method of separation of variables is nothing more than an application of the chain rule, so you can avoid the symbols $dx$ and $dy$ altogether:

If $\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{f(x)}{g(y)}$, then

$\frac{\mathrm{d} y}{\mathrm{d} x}g(y)=f(x)$.

If $g$ has an antiderivative, $G$ then $\frac{\mathrm{d} G}{\mathrm{d} y}=g(y)$ and we get

$\frac{\mathrm{d} y}{\mathrm{d} x}\frac{\mathrm{d} G}{\mathrm{d} y}=f(x)$.

The LHS is $(G\circ y)'(x)$ from which it follows that

$(G\circ y)(x)=\int f(x)dx.\ $ (Note the antiderivative is taken wrt $x$ on $both$ sides).

On the other hand, $(G\circ y)(x)=G(y(x))=G(y)=\int g(y)dy$.

Putting this all together, we have the desired result

$\int g(y)dy=\int f(x)dx.$

Remark: I guess to treat $dx$ and $dy$ rigorously, it is best to regard them as linear functions on tangent spaces.

Matematleta
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