I came across a differential equation:

$$\frac{dy}{dx}=\frac{\sin(\log x)}{\log y}$$. Here is what I tried to do:

I transformed it into this form $$\log y dy=\sin(\log x)dx$$ i.e. $$\int \log y dy=\int \sin(\log x)dx\dots(2)$$ and after that I used integration by parts to finish off the problem.

However,I was told by my teacher that it should instead be $$\int \log y dy=\int \sin(\log x)dx+C$$ where $C$ is a constant of a integration.I argued that the integration had not yet been carried out and so there was no need for the constant.(and I was told it $had$ to be there.)

Can anyone please convince me why my teacher is right and I wrong?



The problem arises because the method of separation of variables is wrongly defined. Here is a bettter way to see the method of separation of variables. Say you have: $$P(x,y)dx + Q(x,y)dy = 0 $$ After algebraic manipulation, if you can put this equation into this form: $$F(x)dx + G(y)dy = 0$$Then you can integrate the equation, and hence you will have $$\int F(x)dx + \int G(y)dy = C $$

  • $\begingroup$ Just wondering would zero be integrated with respect to $x$ or $y$ in this case? I know the answer would probably not change. $\endgroup$ – NoChance Oct 28 '12 at 9:48

My suggestion is that you explicitly write functions with their arguments and avoid shifting around $dx$ and $dy$ as if they were functions or numbers. Of course if you are considering differential forms, these manipulations may be given a very concrete meaning, but otherwise they look like tricks to me. Which are good, but only after you really get the hang of it.

So, you write your equation as:

$$y' ( x) = \frac{\sin ( \log x)}{\log y ( x)}.$$

Where you are probably assuming $y(x)>1$, right? You move it to the left hand side and integrate over a definite interval $[a,\xi]$,with $a$ being the point where you have your initial condition $y(a)=y_a$:

$$\int_a^\xi \log y ( x) y' ( x) dx = \int_a^\xi \sin ( \log x) dx.$$

The left integral is just a change of variable which you can integrate as long as the initial condition is $y_a>1$. You integrate the right integral too and get a function of $\xi$. The constants are now explicit and you can choose to subsume them in some $C$ or just carry them along, but there's no arguing whether there are any or not.

This is not exactly the fastest way, nor maybe the most elegant, but there aren't any ambiguities.

  • $\begingroup$ Unfortunately, we are dealing with differential forms as pointed out here:math.stackexchange.com/questions/221366/… I do not yet understand them(especially the definition given in wikipedia using vector bundles and stuff) but the texts here seem to be using them without pointing out what they really are. $\endgroup$ – user43081 Oct 28 '12 at 12:29
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    $\begingroup$ I guess there are lots of books on the subject, but if you are interested in differential forms in geometry, I can recommend the little book by do Carmo: books.google.de/books/about/… But for applications to ODEs there must surely be a better one. $\endgroup$ – Miguel Nov 5 '12 at 8:01

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