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I have on a textbook a system of two ODEs as follow

$$\dfrac{df}{dx}=af(x)g(x)\\ \dfrac{dg}{dx}=-bg(x)$$

So, the author studies $df/dg$ writing $\dfrac{df}{dg}=-\dfrac{a}{b}f(x)$.

I understand the arithmetic, but I am not sure of mathemathical/geometric meaning.

Thank you in advance for light and maybe some lemmas or theorems of reference that it is valuable (are when is valuable).

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2 Answers

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I have ranted numerous times on this site about that horrendous notation. It really means nothing and can be very misleading. I would suggest you write$$\frac{df}{dx} = af(x)g(x) = -\frac ab f(x)\frac{dg}{dx},$$and then divide by $f(x)$ to get$$\frac1{f(x)}\frac{df}{dx} = -\frac ab \frac{dg}{dx}.$$Now integrate both sides (with respect to $x$) and you get$$\ln |f(x)| = -\frac ab g(x) + c.$$[You could write the previous equation as $\frac{df}f = -\frac ab dg$, and then you get where your textbook should end up.]

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This notation is often difficult to make sense of. What, for instance, would it mean to differentiate $\sin(x)$ with respect to $\cos(x)$? To begin with, we must write $\sin(x)$ in terms of $\cos(x)$:$$ \sin(x)=\sqrt{1-\cos(x)^2} \, , $$ignoring the $\pm$ signs that should be on the square root for now. This means that$$ \frac{d\sin(x)}{d\cos(x)}=\frac{d\sqrt{1-\cos(x)^2}}{d\cos(x)} \, . $$But this is the same as$$ \frac{d}{du}\left(\sqrt{1-u^2}\right)\Biggr|_{u=\cos(x)} \, . $$So every time you see$$ \frac{df(x)}{dg(x)} \, , $$note that $f(x)=h(g(x))$ for some function $h$, and write$$ \frac{dh(g(x))}{dg(x)}=\frac{dh(u)}{du}\Biggr|_{u=g(x)} \, . $$The $u$ is a dummy variable that can then be replaced by whatever function you are 'differentiating with respect to'. You can use $g(x)$ as a dummy variable, but it often leads to confusion, and is considered by many people to be poor style.

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To give another example, let's consider the chain rule,$$ \frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx} $$Say $y=\sin(u)$, where $u=x^2$. Then, the chain rule appears to say$$ \frac{d}{dx}(\sin(x^2))=\frac{d}{dx^2}\left(\sin(x^2)\right) \cdot \frac{d}{dx}(2x) \, . $$It seems difficult to parse the first term on the RHS. The easiest way to do this is to write it instead as$$ \frac{d}{du}(\sin(u)) \Biggr|_{u=x^2} \, . $$However, we can also treat $x^2$ just as we would any other variable, and make sense of the term that way, too.

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