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It is a common practice to move the $dx$ around when solving ODE and we take for granted when we integrate both sides. However, I've been rather uncomfortable with this. From an analysis perspective, $dx$ itself doesn't make sense to me. Instead, we always consider $\frac{df(x)}{dx}$, or the operator $\frac{d}{dx}$ by itself.

Now, some people argue that both $dy$ and $dx$ are infinitesimal things that are not equal to zero, and thus we can move them around. But that doesn't convince me neither. Shouldn't we concretely define what $dy$ and $dx$ are before we do anything to them? Is it possible to justify moving $dx$ around by using $\delta-\epsilon$ analysis?

As for integrating both sides, what annoys me is that sometimes extra terms (usually the constant $C$) appear. It's tempting to think of "integrating both sides" as a kind of operations (maybe not binary?) just like add or subtract, but that doesn't seem right because by apply indefinite integral a function we actually get a bunch of functions. What is the right way to think of this process? Should we think about this process under the framework of algebra, maybe?

Hope my questions make sense. Thanks.

Viktor Vaughn
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    Possible duplicate of this thread: http://math.stackexchange.com/questions/47092/physicists-not-mathematicians-can-multiply-both-sides-with-dx-why – Quinn Greicius Feb 27 '16 at 04:10
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    From differential topology point of view $dx$ is differential 1-form. Integrating 1-forms makes perfect sense. There is also a point of view from non-standard analysis. – Predrag Punosevac Feb 27 '16 at 04:14
  • http://math.stackexchange.com/questions/852394/failure-of-differential-notation/1651264#1651264 –  Feb 27 '16 at 04:30
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    I am sure there are more rigorous ways to do it, but try "pretending" that $dx$ is just a "very very small number" and then it becomes easier to understand. – MathematicsStudent1122 Feb 27 '16 at 04:38
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    The simplest not-hand-waving proof that I know treats this moment following way. If you have a differential equation $y' = \frac{P(x)}{Q(y)}$ then it's true that $Q(y) y' = P(x)$. If you plug a solution into this formula, you get functions of $x$ at both sides of equality. You can integrate now with respect to $x$ (using definite or indefinite integration). But then you see that $\int Q(y(x)) y'(x) , dx $ reminds you formula for change of variables under the integral and it becomes $\int Q(y) , dy$. For this kind of thinking multiplying by $dx$ is just a shortcut. – Evgeny Feb 27 '16 at 06:56
  • When undergrad differential equations textbooks manipulate $dx$ and $dy$ as if they were independent quantities, you can always easily rephrase the argument to avoid doing so. In that way a nonrigorous argument can be converted into a rigorous argument. – littleO Dec 25 '22 at 04:56
  • Is there some way to add this (somewhere) as a faq? – A rural reader Feb 26 '24 at 03:37

1 Answers1

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Answering your latter question... Integrating both sides does preserve equality. The arbitrary constant encountered in an indefinite integral still appears in the solution to a differential equation that has unspecified initial conditions. Consider a simple differential equation:

$$\begin{align} \frac{dy}{dx} &= 2xy \\ \int \frac{dy}{y} &= 2\int x\ dx \\ \ln|y| +C_1 &= x^2 + C_2 \\ \ln|y| &= x^2 + C \quad\quad\text{where $C=C_2-C_1$} \\ |y| &= e^{x^2 + C} \\ |y| &= e^{C}e^{x^2} \\ |y| &= C e^{x^2} \quad\quad\text{since $C$ and $e^C$ are both arbitrary constants} \\ y &= C e^{x^2} \quad\quad\text{again $C$ could be either positive or negative. It's undetermined.} \end{align}$$

We now have the general solution the differential equation. It is a family of solutions that depends on some given condition. If we did have an initial condition, rest assured that we would then find a unique solution to our ODE.

Gary
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