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So im reading a book called "Ordinary Differential Equations" (Tenenbaum & Pollard) and in the introduction(ish) they are doing an example using a carbon dating problem, represented as:

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

Which they change to

$\frac{dx}{x} = -k dt$

And then integrate to

$ \log x = -kt + c$

But doesn't that imply integrating with respect to x in the left side and integrating with respect to t on the right? Is there a proof that says that is okay? I would imagine it is related to the Fundamental Theorem of Calculus, but i didnt see anything.

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    $\begingroup$ This is the chain rule in disguise! Used to bother me as well $\endgroup$
    – operatorerror
    May 14, 2016 at 23:58

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This is a stealthy application of the chain rule. See here for an explanation:

http://www.math-cs.gordon.edu/courses/ma225/handouts/sepvar.pdf

For convenience of future users I will summarize the argument here:

Suppose we have some differential equation $g(x)\frac{dx}{dt}=f(t)$ where we are trying to recover a function $x(t)$:

Then we can certainly integrate with respect to $t$ on both sides: $\int g(x)\frac{dx}{dt}dt=\int f(t)dt$. But since $x(t)$ is a function of $t$, we have, by the chain rule and setting $G(x)$ s.t. $G'(x)=g(x)$ $\int g(x)\frac{dx}{dt}dt=G(x(t))=\int f(t)dt$, which you can check by differentiating.

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  • $\begingroup$ Are you assuming that the $$\frac{dt}{dt}$$ cancels out? Thanks $\endgroup$
    – information_interchange
    Jan 24, 2020 at 21:01

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