I have started reading a book on differential equations and it says something like:

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

Integrating both sides gives

$$\log x = kt + c$$

How is it that I can 'integrate both sides here' when I am integrating one side with respect to $x$ yet I am integrating the other side with respect to $t$?

  • 1
    $\begingroup$ When you write "k dt" in $\TeX$, with a space between "k" and "dt", no space appears in the final product; you get $k dt$. But if you write "k \, dt", then you get $k \, dt$. That's how I and many others do this. I've edited the question accordingly. $\endgroup$ – Michael Hardy Jun 22 '12 at 22:30

What is going on there is what is called an abuse of notation. What you really have there is an equation in $t$. $x=x(t)$ is a function dependent on $t$. So what we're doing is the following - I presume this is the original equation:

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

This is the same as

$$x'(t)=k x(t) $$

$$\frac{x'(t)}{x(t)}=k $$

Now we integrate wrt to $t$

$$\int\frac{x'(t)}{x(t)}dt=kt+C $$

But we note letting $X=x(t)$ so $dX = x'(t) dt$ gives

$$\int\frac{dX}{X}=kt+C $$

$$\log X = kt+C$$

So switching back

$$\log x(t) = kt+C$$

$$x(t)=C e^{kt}$$

What we actually do, in some sense, is integrate with respect to "only" $t$ in one side, and "$x(t)$" in the other (which is done implicitly). The notation is very useful and suggestive, so we use it, understanding what we're doing is the above.

  • $\begingroup$ I'm not so sure this is an abuse of notation. It's an abuse of notation if one takes conventional rigorous definitions literally. But sometimes that only means the rigorous definitions that have been developed so far have not yet caught up with what can be done. $\endgroup$ – Michael Hardy Jun 22 '12 at 22:28
  • $\begingroup$ @MichaelHardy Regardless, it is, in my opinion. $\endgroup$ – Pedro Tamaroff Jun 22 '12 at 22:30
  • $\begingroup$ "In mathematics, abuse of notation occurs when an author uses a mathematical notation in a way that is not formally correct but that seems likely to simplify the exposition or suggest the correct intuition (while being unlikely to introduce errors or cause confusion)" I mostly agree with that. $\endgroup$ – Pedro Tamaroff Jun 22 '12 at 22:31
  • $\begingroup$ Whom are you quoting? And what does "formally" mean? Formally often means something fits a form but is not logically sound; e.g. there are "formal power series" that do not converge. To apply rules of algebra "formally" is to apply them beyond the context in which one knows them to be correct if taken literally. Of course that's not what was meant in the quote, but maybe a better choice of word would be possible. $\endgroup$ – Michael Hardy Jun 22 '12 at 22:53
  • 1
    $\begingroup$ Oh wait, I get it, C just stands for the arbitrary constant $e^C$ $\endgroup$ – Jim_CS Jun 23 '12 at 9:00

Good point! We have $\frac{dx}{dt}=kx$. Thus $x$ is a function of $t$. Crossing our fingers about possible division by $0$, we have $$\frac{1}{x}\frac{dx}{dt}=k.\tag{$1$}$$ Now let $F(x)$ be any antiderivative of $\frac{1}{x}$ with respect to $x$. Then by the Chain Rule, the left-hand side of $(1)$ is the derivative of $F(x)$ with respect to $t$. So we have $$\frac{d}{dt}(F(x))=k,$$ and therefore $F(x)=kt+C.$ In our case, $\log(|x|)$ is an antiderivative of $\frac{1}{x}$ with respect to $x$, and we get the general solution of the DE.

Note that the mysterious process in which we separate $\frac{dx}{dt}$, which is not a ratio, into $2$ parts, "$dx$" and "$dt$," gives us exactly the same final answer. We can think of it as a symbolic manipulation that gets us to the right answer.

Remarks: $1.$ Note that $x=0$ is a solution of the DE. If we use $\log(|x|)=kt+C$, then $x=\pm e^C e^{kt}$, and $e^C$ can never be $0$. However, it is traditional to replace the constant $\pm e^C$ by a new constant $D$, so we get $x=De^{kt}$. The case $D=0$ covers the solution $x=0$ that we had lost by dividing. A nice case of two mistakes cancelling.

$2.$ At a (much) later stage, the "differentials" that we used can be given precise meaning.


Letting $y= \log x$, we have $$ \frac{dy}{dx} = \frac 1 x. $$ If we regard $dy$ and $dx$ as corresponding infinitely small increments of $y$ and $x$, and multiply both sides by $dx$, we find that the infinitely small increment of $y$ is $$ dy = \frac{dx}{x}. $$ If $y=kt+c$, and we increment $y$ and $t$ by infinitely small amounts while letting $k$ and $c$ remain constant, we get $$ dy = k\,dt. $$ Thus $$ \frac{dx}{x} = k\,dt,\text{ and }\log x = kt+c. $$

One side is integrated with respect to $x$ and the other with respect to $t$ because $dx$ appears as a factor (i.e. as something one multiplies by) on one side and $dt$ on the other.

If anyone finds deficiencies in the extent to which this has been made logically rigorous in the existing literature, I think that should be construed as a challenge to work further on the logic and the conventions so that the above will be fully logically rigorous. What appears above should not be viewed simply as an abuse of notation. It's an abuse of notation only within the context of certain conventions, which may represent the state of the science at some particular point in history. The current state of the science is never the last word.

  • 2
    $\begingroup$ Michael, I think we have left infinitesimals out of the picture of standard analysis, but non-standard is there (which I know very little about). What I think is that this is being asked in terms of standard analysis, in which it is the case it is an abuse of notation. Glad to disagree, though. $\endgroup$ – Pedro Tamaroff Jun 22 '12 at 22:52
  • $\begingroup$ And there is also "smooth infinitesimal analysis". What I wrote above comes closer to being logically rigorous in smooth infinitesimal analysis, with its nilpotent infinitesimals, than in non-standard analysis. In the latter, one would probably at some point talk about "standard parts", etc. $\endgroup$ – Michael Hardy Jun 22 '12 at 22:55
  • $\begingroup$ Oh, good to know. $\endgroup$ – Pedro Tamaroff Jun 22 '12 at 22:57
  • $\begingroup$ @PeterTamaroff : One achievement of non-standard analysis that seems not to have been anticipated (but I could have missed it) by the sort of informal reasoning with infinitesimals done by Euler et al., is the way NSA distinguishes between continuity and uniform continuity. If $f(x+dx)-f(x)$ is infinitesimal whenever $x$ is real and $dx$ is infinitesimal, then $f$ is continuous. But if that persists not only when $x$ is real, but also when $x$ is a non-standard real, then $f$ is uniformly continuous. Thus with $f(x)=\sin(1/x)$, if $x$ is infinitesimal, then [.....] $\endgroup$ – Michael Hardy Jun 22 '12 at 23:17
  • $\begingroup$ $f(x)$ can go all the way from $-1$ to $1$ while $x$ changes by an infinitely small amount, so we don't have uniform continuity. And if $x$ is infinitely large, the $e^x$ can change by more than an infinitesimal as $x$ changes by an infinitesimal, so again there's a lack of uniform continuity. $\endgroup$ – Michael Hardy Jun 22 '12 at 23:18

You could also view integration as a linear operator you apply to both sides of the equation. Applying the same operator to both sides still gives you a valid equation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.