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Suppose we have the differential equation which can be expressed symbolically as $$ \frac {dx} x = -k \ dt$$ where $x$ is a function of $t$.

Now, my book says we can take the antiderivative of both sides to get

$$\log x = -kt + C$$

and I understand this. Now let's take a step back: The original diff eq above can also be written (maybe more commonly) as

$$\frac {dx}{dt} = -kx$$

which contains the exact same information, I think.

Since it's informationally the same as the first equation, we ought to be able to take the antiderivative of the above equation, just like as we did with the first equation.

Yet taking the antiderivative in the above form seems problematic since there are two differentials on the same side (raising the question "What would the antiderivative be taken with respect to?")

My question to you is "How is this conventionally treated?" i.e.

How is "taking the antiderivative of an expression" defined in the case when there is more than one differential on a given side of the equation?

EDIT: I know this question is different from the question in my title. (In fact it is a special case of it, so providing a good answer to either the question in the title or the bolded question above would be greatly appreciated.)

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    I think you're asking the wrong questions here. It seems to me that the real questions are: what are $dx$ and $dt$ in $\frac{dx}{x}=-k dt$ and what are the antiderivatives of both sides of this equality? The answers are that $\frac{dx}{x}=-k dt$ is a heuristic way to rewrite $\frac{dx}{dt}=-kx$ where variables have been separated and that the fact that you can write $\int\frac{1}{x}\,dx=\int -k\,dt$ in order to obtain $x(t)$, while correct, is merely a shortcut way of proceeding that can be rigorously proved to work. – 2017-02-20
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    Thanks, this is helpful - please elaborate. What are $dx$ and $dt$, and how would such a rigorous proof work? – 2017-02-20
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    Starting from $\frac{dx}{dt} = -kx$ (where $\frac{dx}{dt}$ is not a fraction of differentials, but instead **the derivative**), divide by $x$ on both sides and then antidifferentiate with respect to $t$: $$\int \frac 1x\frac{dx}{dt}dt = \int -kdt$$ You get exactly the same result (remember to use the FTC/ substitution on the LHS to rewrite it as $\int \frac 1xdx$). – 2017-02-20
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    Thanks! I am wondering, why does the FTC allow us to make the substitution you mentioned? Also, why is it necessary to get the $x$ symbols all on one side before antidifferentiating with respect to $t$? – 2017-02-20
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    @Asker You could use FTC or substitution to show that $\int \frac 1x\frac{dx}{dt}dt = \int \frac 1xdx$. But substitution is easier. It's impossible to evaluate $\int -kxdt$ where $x$ is an arbitrary function of $t$. If it's a given function, we *might* be able to do it. But $\int \frac 1x\frac{dx}{dt}dt$ is not only possible to evaluate even when $x$ is some arbitrary (meaning unknown) differentiable function of $t$, it's downright easy. – 2017-02-20

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