\[\int \frac{dx}{lnx}\]

- anonymous

\[\int \frac{dx}{lnx}\]

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- schrodinger

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- anonymous

How to start?

- anonymous

Do you know how to use integration by parts?

- anonymous

Integration by parts doesn't look nice..

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## More answers

- anonymous

\[\int \frac{dx}{lnx} = \frac{x}{lnx} - \int x d(\frac{1}{lnx})\]If that's what you meant...

- amistre64

u=lnx; x=e^u
du=dx/x
x du = dx
\[\int \frac{1}{u}e^u~du\]
maybe?

- amistre64

http://www.wolframalpha.com/input/?i=integrate+1%2Flnx
you sure you got it right?

- anonymous

\[\int \frac{dx}{lnx}\]\[=\int \frac{x}{xlnx}dx\]\[=\int \frac{x}{lnx}d(lnx)\]\[=\int \frac{e^{lnx}}{lnx}d(lnx) \]\[=\int \frac{e^u}{u}du\]\[=\int \frac{1}{u}d(e^u)\]....
The original question is
\[y' = \frac{1}{y^2lnx}\]\[y^2 dy = \frac{dx}{lnx}\]I supposed integrating both sides can give me the answer :/

- anonymous

Perhaps I'm in the wrong direction?!

- amistre64

it looks like any solution will have to run into a logramathic integral

- anonymous

What technique(s) is(are) required to solve this equation?

- amistre64

ones that i havent run into yet :/

- amistre64

you can try to develop a series poly for it, and see what that would bring you

- anonymous

:O I haven't learnt expanding the function!

- amistre64

what methods have you learnt?

- anonymous

separable, integrating factors, exact question, undetermined coefficients, variation of parameters, homogeneous equation...
Btw, I just ran around and grabbed the exercise, I think it's not in my textbook :/
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/readings/notes_exe/MIT18_03S10_1ex.pdf
1A-3a

- anonymous

*exact equation lol

- amistre64

y^2 y' = 0 when y'=0 or y^2=0; given y1=0 and y2=C
not sure of a wronskian would be useful here or not

- anonymous

Actually, how does Wronskian work. I just know it can determine the linear dependency.. Then, nothing :/

- amistre64

the wronskian is the short version of the undetermined coeffs

- amistre64

Wx W Wy
f1 0 f2
f'1 g f'2
then its similar to the cross product
W = f1f'2 - f'1f2
Wx = -g f2
Wy = f1 g
and you integrate Wx/W and Wy/W

- amistre64

yeah, i dont see it being useful here tho

- anonymous

It...somehow... reminds me of Cramer's :/

- amistre64

it is, since the cramer is what you do at the end of the undetermined coeffs

- anonymous

Ahhhh.... Thanks for teaching something new though!!!

- anonymous

*teaching me something

- amistre64

yep, but as far as a solution to this goes; i got no idea. The wolf says we need a logartithmic integral which leads me to believe that this might be workable with a power series solution; but i dont think i can think thru it this late

- anonymous

Never mind, perhaps i should ask my teacher after the lesson today.
I have to go for lesson now. Once again, thanks for your help!!

- anonymous

Sorry for the very late reply. I asked my teacher some days ago and he said I wouldn't know how to do it. Even for some graduate student, they didn't know that too. And he didn't teach me how to do the integration....
Thanks for your time helping me out!!

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