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For example, to solve $$\frac{dy}{dx} = \frac{x-5}{y^2}$$

we turn the equation into

$$y^2dy = (x-5)dx$$

Even though $\frac{dy}{dx} = \frac{d}{dx}(y)$ is not a fraction. The $\frac{d}{dx}$ notation denotes an unary operation (I think).

This abuse of notation has been discussed extensively on here, but I haven't seen any posts explaining why it works in the case of the separation technique of solving differential equations. Can anyone explain/point me to an appropriate resource?

James Ronald
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    Does this help? https://math.stackexchange.com/questions/27425/what-am-i-doing-when-i-separate-the-variables-of-a-differential-equation – teddy May 14 '20 at 00:37
  • What so you mean by saying $\mathrm dy/\mathrm dx$ is no fraction? Of course it is a fraction. There is no abuse of notation anywhere. The symbols $\mathrm dy$ and $\mathrm dx$ denote differentials. – Allawonder May 14 '20 at 15:22

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Because you are actually implicitly treating $x$ and $y$ as functions of another variable, perhaps "$t$". Somewhere, you should have been exposed to the fact that the slope of the parametric curve $$ c(t) = (x(t), y(t) $$ is $$ \frac{y'(t)}{x'(t)} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t} \text{.} $$ Then you are syntactically manipulating "$x$", "$y$", "$\mathrm{d}x$", and "$\mathrm{d}y$" as placeholders for $x(t)$, $y(t)$, $\mathrm{d}x/\mathrm{d}t$, and $\mathrm{d}y/\mathrm{d}t$, respectively.

Perhaps surprisingly, you have been doing exactly this since you first studied implicit differentiation in Calculus.

Eric Towers
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If your equation has the form $$ y'=M(x)N(y) $$ you call $\Phi(y)=1/N(y)$ where $N\neq 0$. Then your equation reads $$ \Phi (y(x))y'(x)=M(x) $$ which should be an identity over some interval. So you can integrate in $x$ over any interval and get an identity, typically you choose the the lower limit to be the point $x_0$ where you know your "initial" data $y_0=y(x_0)$. But you can change the variable on the left integral and integrate in $y$ instead $$ \int\Phi(y)\,dy=\int M(x)\,dx $$ which is exactly what you do when you separate the variables.

BTW, $dy/dx$ was a fraction for Leibnitz, for Euler and Lagrange, and for any Physicist, Engineer or person with common sense. Over the last 100 year or so Mathematicians became picky about this and sacrificed the beauty of Leibnitz' notation on the altar of rigor ("rigor mortis").

GReyes
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  • just a comment: before mathematicians "became picky", half of analysis was wrong (not just imprecise, actually flat out wrong). example: until Weierstrass gave a counterexample, it was widely assumed that any function was differentiable outside countable sets, or that limits of differentiable functions were differentiable (without of course being too precise on the concept of the limit of a sequence of functions). – Albert May 14 '20 at 14:18
  • @Glougloubarbaki You are right. However, there is much more to be said here. Being right or wrong is just one aspect of it. There is also the pedagogical aspect. Calculus (and, by extension, Analysis) was born from the consideration of changing quantities that appear in Mechanics, Thermodynamics, etc. Infinite and infinitesimal quantities are at the very heart of Calculus. I strongly believe that the algebraization of Calculus, whereby those are replaced by limits and derivatives are no longer quotients of infinitesimals, have been nefarious for its teaching. – GReyes May 14 '20 at 16:39
  • I have experienced how painful and useless it is for students to go through the epsilon-delta formalism. I knew an Engineer who told me that he had to teach himself useful Calculus because the one he was taught by Mathematicians was perfectly rigorous and useless. I would rather use Euler's "Introductio" to teach Calculus than the horrible textbooks that we use now, deprived of any soul or interesting application, with a whole chapter on epsilons and deltas. – GReyes May 14 '20 at 16:44
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The "abuse of notation" is a usefull shortcut to avoid a lot of intermediate steps. If not : $$\left(\frac{dy}{dx}\right) = \frac{x-5}{y^2}\qquad y\neq 0$$ $$y^2\left(\frac{dy}{dx}\right) = y^2\frac{x-5}{y^2}$$ $$y^2\left(\frac{dy}{dx}\right) = x-5$$ $\frac13\frac{d(y^3)}{dx}=y^2\left(\frac{dy}{dx}\right)$ $$\frac13\frac{d(y^3)}{dx}=x-5$$ Without multiplication by $dx$ but from definition of antiderivative : $$\frac13 (y^3)=\int(x-5)dx=\frac12 x^2 -5x+c$$ Note : Nowadays this "abuse of notation" is no longer an abuse of notation. It is fully justified thanks to the Non Standard Analysis theory. But that's another story.

JJacquelin
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