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Today, for the first time, I touched on the topic of "separation of variables" and my mind was blown because I did not know that some actions that we take in relation to the manipulation of $\frac{dy}{dx}$ were possible.

For example,

$$\frac{dy}{dx} = - \frac{x}{1+y^2}$$

To solve this, my textbook literally took the $dx$ and shifted in on the other by multiplying it and I was like "how is that possible?!". I thought derivatives were fixed (you cannot meddle with them as you meddle with fractions using rules of algebra).

Not only that the second step they did was take the integral of both sides. Like, I mean, is that even valid algebraically? I have never seen that being done before :O

Please, could someone please clarify this, maybe a proof or so? Thanks in advance.

Vítězslav Štembera
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Shooting Stars
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1 Answers1

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$\newcommand{\d}{\mathrm{d}}$This has been asked before but I’ll add a simpler observation. A fully rigorous justification for “the physicist’s method” requires an understanding of differential forms, but in the problem you are stating we don’t need that - and I don’t understand them myself. You can research that yourself - be warned it’s much more advanced than separation of variables - but I’ll just show you that that level of formalism isn’t necessary for a lot of cases.

For many separation-of-variables questions, it is essentially a different way of doing the following perfectly valid manipulation:

$$\begin{align}\frac{\d y}{\d x}=-\frac{x}{y^2+1}&\implies(y^2+1)\cdot\frac{\d y}{\d x}=-x\\&\implies\int(y^2+1)\cdot\frac{\d y}{\d x}\,\d x=-\int x\,\d x\\&\implies \frac{1}{3}y^3+y+c_1=-\frac{1}{2}x^2+c_2\\&\implies\cdots\end{align}$$

And we used the fact that the chain rule states: $$\frac{\d}{\d x}\left[\frac{1}{3}y^3+y+c_1\right]=[y^2+1]\cdot\color{red}{\frac{\d y}{\d x}}$$

And this is what you would get if you more lazily “multiplied by $\d x$”.

FShrike
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