# Does multiplying by $dt$ have any meaning?

Consider, for example, the equation $x'=x$, then it is usually solved by writing $\frac{dx}{dt}=x\implies\frac{dx}{x}=dt\implies\int\frac{dx}{x}=\int dt$ ...

I know that there is a theorem in ODE that justify $x'=x\implies\int\frac{dx}{x}=\int dt$ ,but my question is about the intermediate step: $x'$ at some point $x_{0}$ is defined via a limit. $\frac{dx}{dt}$ is, as far as I understand, a notation for the function $x'$ - so we can not multiply by $dt$ since it has no meaning, it is a part of the notation.

My question is as follows: Although the last step is indeed correct and can be justified, does the intermediate step (multiplying by $dt$) have any meaning, or is it just an easy way to remember and get to the last step ?

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It can be formalized, i.e. via nonstandard analysis. One can regard $dt$ as a variable in its own right and manipulate it under certain circumstances. – Alex Becker Jul 25 '12 at 1:44
You may be interested in en.wikipedia.org/wiki/Differential_of_a_function – Argon Jul 25 '12 at 1:45
This has been asked several times here... There was one such question just one or two days ago. – Hans Lundmark Jul 25 '12 at 8:24

It's really a formalism supported by the fundamental theorem of calculus and the chain rule.

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I don't think it's ONLY a formalism supported that way. There are things in mathematics that nobody understands, but that are nonetheless real. People try to make this rigorous via nonstandard analysis or via smooth infinitesimal analysis, but I think methods like those have captured only part of what's there. – Michael Hardy Jan 23 '13 at 3:51

I did a rather extensive write-up on this in another discussion, so as not to duplicate things too much I'll post a link to that: Link. To summarize:

Starting with a seprable ODE: $${{dx(t)} \over {dt}} dt = g(t)$$

The process of "multiplying by the differential" is executing an integration by substitution using the function f(x) as the substitution variable, so:

$$u = x(t)$$ $$du = x'(t)dx$$ $$\int {{dx(t)} \over {dt}} dt = \int du = \int dx(t) = \int g(t) dt$$ which results in: $$x(t) = \int g(t) dt$$

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You've got a pretty bad typo above. You should have $\dfrac{dx(t)}{dx}\,dt = g(t)\,dt$. – Michael Hardy Jan 23 '13 at 3:52