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I know a number of ways of solving this basic DE:

$\ddot{u} = -u$

Besides the fact that the solution is obvious, one can do:

$\ddot{u} = \frac{d\dot{u}}{dt} = \frac{d\dot{u}}{du}\frac{du}{dt} = \frac{d\dot{u}}{du}\dot{u} = -u$

and then bring the du in the denominator to the right and integrate both sides to get:

$\dot{u}^2 = - u^2$ or $\dot{u} = \pm i u$

Then if the solution still wasn't obvious to us, we could separate variables again to get u(t) = e$^{\pm it}$

Now my question is, why is it not possible to use separation of variables more directly, that is, what is wrong with the following:

$\ddot{u} = \frac{d}{dt}\frac{du}{dt} = - u$

and bring the dt over to the right and u to the left to get:

$\frac{d}{dt}\frac{du}{u} = - dt$

Which can be integrated on both sides. The problem is that this doesn't seem to work, you get a wrong solution to the DE (u(t) = e$^{-t^2/2}$), even if you keep careful track of the limits of integration. But I can see no problem with it; the solution we know is well-behaved, doesn't have any singularities and is differentiable. What goes wrong?

user1247
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  • ... bring the $dt$ over to the right and $u$ to the left is wrong. – Quang Hoang Aug 19 '14 at 14:32
  • @Quang Hoang, without more explanation that is not helpful, because that is exactly what you do in textbook separation of variables. – user1247 Aug 19 '14 at 14:48
  • Loosely speaking, when you perform separation of variables, you only swap the outer $dt$. Look at the equation you got after your operation: LHS is a function but RHS is a differential. – Quang Hoang Aug 19 '14 at 14:56
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    @user1247: No, that is not at all what you do in the usual case, because then there is no $d/dt$ "standing in the way". You can't bring $dt$ and $u$ past the differential operator $d/dt$ just like that. (Anyway, the usual way of doing separation of variables is just a mnemonic for the chain rule, so it should perhaps not be taken too literally, but that has already been discussed many times on this site.) – Hans Lundmark Aug 19 '14 at 15:06
  • @Hans Lundmark, you almost always can if you are dealing with well-behaved functions (as we are here). I am a physicist (which should tell you something) and ignoring the order of differentiation is practically assumed for every function we tend to work with. So why doesn't it work here? BTW I searched the site for the same question and couldn't find it answered before. – user1247 Aug 19 '14 at 16:54
  • Well, no. This is not about equality of mixed derivatives, it's about the fact that the operation "multiplication by $u(t)$" followed by the operation "differentiation with respect to $t$" is not the same thing as doing those operations in reverse order. (A very fundamental fact in quantum mechanics, if you want to talk physics.) – Hans Lundmark Aug 19 '14 at 22:26
  • @Hans Lundmark, it's not obvious to me at least that that is what is being done here. When I think of the "du" and "dt" I am thinking of infinitesimals (I think that term is right?) which I thought had a rigorous basis. If I just write down the relevant line using limits to define what I mean by differentiation, then there are no operators, and so I can't find where I am doing anything wrong. It would help if you were more explicit about exactly when this kind of thinking (either using infinitesimals or limits to work with only c-values) breaks down. – user1247 Aug 19 '14 at 22:56
  • When you bring $u$ down on the left-hand side you get $\frac{1}{u} \frac{d}{dt}\left(\frac{du}{dt} \right)$, and this is definitely not equal to $ \frac{d}{dt}\left(\frac{1}{u} \frac{du}{dt} \right)$ (for the reason that I explained: $d/dt$ doesn't commute with multiplication by $1/u$). And what it would actually mean to take the inner $dt$ outside $\frac{d}{dt}$ and up on the right hand side, I can't even begin to imagine... I just doesn't make sense in any way. – Hans Lundmark Aug 20 '14 at 00:10
  • @Hans Lundmark, thanks. I do understand this, I just don't understand exactly when the heuristic use of infinitesimals breaks down. I guess you can only reliably use them in first order DEs for the reason you gave. – user1247 Aug 20 '14 at 13:43
  • Yes. As I said, the usual recipe for separable ODEs is a mnemonic for that particular situation, and it works because it is the chain rule in disguise: http://math.stackexchange.com/a/27433/1242 – Hans Lundmark Aug 22 '14 at 09:41

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$$\frac{d\dot u}{du}$$ don't give you anything because $\dot u$ is not dependent of $u$ (in other word, it's not of the forme $\dot u(u(t))$).

Here, you have to do as usual: solution are of the forme $$A\cos(t)+B\sin(t).$$

idm
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