I have asked questions before, and have read many questions and answers from others on stackexchange, about how to treat "$dx$" in a differential equation.

people often have the intuition of treating $dx$ in the term "$\frac{df}{dx}$" as if it were a separate subterm of that, even though $\frac{df}{dx}$, or at least $\frac{d}{dx}$ is an operator, and shouldn't be seen as a fraction of two seperable elements.

Nevertheless, when solving the simple differential equation $\frac{dx}{dt}=x$, people often proceed as follows:

$$(a):\qquad \frac{dx}{dt}=x$$ Step 1: multiply by $dt$ and subtract by $x$: $$(b):\qquad \frac{1}{x}dx=dt $$ Step 2: integrate both sides: $$(c): \qquad \int\frac{1}{x}dx=\int dt$$ Step 3: Solve the integral: $$(d): \qquad ln(x)=t+C\implies x=e^{t+C}=x_0e^t$$

So my questions are

  1. we know that the justification of Step 1 as "multiplication" is incorrect, since $dx$ and $dt$ are not seperable elements, so what is the justification for going from equation $(a)$ to $(c)$?

  2. When $a=b$, we can conclude that $\int adx=\int bdx$, since this takes the integral of both sides with respect to the same variable, but What is the justification of inserting the integral sign in equation $(c)$, without also adding the differential $dx$ to both sides?


marked as duplicate by MrYouMath, Hans Lundmark, Community Sep 29 '16 at 14:25

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  • $\begingroup$ From (a) to (c) you can do standard integration by substitution: $\int \frac1x \frac{dx}{dt} dt = \int \frac 1x dx$. How that is proven varies from book to book, but it can be done without interpreting $\frac{dx}{dt}$ as a fraction. I don't even know (b) means in a normal calculus class. Differential geometry, sure, but not calculus, or even analysis. $\endgroup$ – Arthur Sep 29 '16 at 11:58
  • 1
    $\begingroup$ @Programmer2134: See near bottom of page 1: math.dartmouth.edu/~m3cod/klbookLectures/303unit/sep.pdf $\endgroup$ – Moo Sep 29 '16 at 11:58
  • $\begingroup$ @Moo, Thanks that was a really good reference. $\endgroup$ – user56834 Sep 29 '16 at 14:25

The proof hops over (b) and goes straight from (a) to (c).

Generally speaking we have (if $f(x)\not=0$): $$\dfrac{dx}{dt}=f(x)g(t)\Leftrightarrow \int\dfrac{1}{f(x)}dx = \int g(t)\,dt$$ since differentiating the RHS with respect to $t$ gives $$\dfrac{1}{f(t)}\dfrac{dx}{dt}=g(t)$$ according to the chain rule.


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