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By long "s", I mean this $\int$

What I am referring to is when we use separation of variables in solving differential equations;

Eg:

  • $\frac{dy}{dx} = \frac{y}{x-3}$
  • $\frac{1}{y} dy = \frac{1}{x-3} dx$

Since this is my first time learning about the separation of variables, I got stuck here. In an example I read earlier, they just put the long "s" in to make it an integral. So my question is, an integral without the long "s" is still an integral? I thought that already needed to be present to compute the integral.

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2 Answers 2

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Formally speaking, the symbols $dx$ and $dy$ have no meaning in isolation, just as $x\to$ has no meaning, except in the context of $\lim_{x\to a}f(x)$. Therefore, while it makes perfect sense to write $$ \frac{dy}{dx}=\frac{y}{x-3} \, , $$ it doesn't really make any sense to write $$ \frac{1}{y} \, dy=\frac{1}{x-3} \, dx \, . $$ So to answer your question directly, expressions such as $$ \frac{1}{y} \, dy $$ have no meaning at all in most treatments of calculus; they are certainly not integrals.

Nevertheless, it can be proven that "separating the variables" and appending an integral sign to both sides of the equation does produce the correct result to these kinds of differential equations. So it is okay to use this method, but it is best to think of it as symbol pushing that happens to produce the correct answer, rather than a rigorous mathematical argument.

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    $\begingroup$ To add to the confusion, exact differential equations are a thing and are related. But your argument stands, it is better to consider this calculus as an informal application of the chain rule or the associated substitution rule of integration (just as product rule of differentiation and integration-by-parts are associated). $\endgroup$ May 26 at 10:07
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    $\begingroup$ Once you learn about differential geometry and differential forms, it does make sense to write something like $\omega = dy/y - dx/(x-3)$, which is a differential form of degree 1. Solving the original differential equation amounts to find a function $f$ such that the pullback of $\omega$ along $\gamma(t) = (t, f(t))$ vanishes. But this is a lot of machinery that is completely useless to an undergrad trying to solve a differential equation. And things get hairier when the DE is nonlinear in $y'$. $\endgroup$ May 26 at 19:54
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Without delving into differential operators, yes. The integral symbol is what makes it an integral. Think of the integral symbol as an operator acting on whatever you are integrating. And in terms of separation of variables not being rigorous, think of it like this: $$\frac1y\frac{dy}{dx}=\frac1{x-3} \\\int\,dx\frac1y\frac{dy}{dx}=\int dx\frac{1}{x-3} \\\int\frac{dy}{y}=\int\frac{dx}{x-3}$$ In this sense, you can see we are using equal operators on both sides to conserve the equality

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