Question

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

Answer 1

How to start?

Answer 2

Do you know how to use integration by parts?

Answer 3

Integration by parts doesn't look nice..

Answer 4

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

Answer 5

u=lnx; x=e^u

du=dx/x

x du = dx

\[\int \frac{1}{u}e^u~du\]

maybe?

Answer 6

http://www.wolframalpha.com/input/?i=integrate+1%2Flnx

you sure you got it right?

Answer 7

\[\int \frac{dx}{lnx}\]\[=\int \frac{x}{xlnx}dx\]\[=\int \frac{x}{lnx}d(lnx)\]\[=\int \frac{e^{lnx}}{lnx}d(lnx) \]\[=\int \frac{e^u}{u}du\]\[=\int \frac{1}{u}d(e^u)\]....

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 :/

Answer 8

Perhaps I'm in the wrong direction?!

Answer 9

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

Answer 10

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

Answer 11

ones that i havent run into yet :/

Answer 12

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

Answer 13

:O I haven't learnt expanding the function!

Answer 14

what methods have you learnt?

Answer 15

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

Answer 16

*exact equation lol

Answer 17

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

Answer 18

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

Answer 19

the wronskian is the short version of the undetermined coeffs

Answer 20

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

Answer 21

yeah, i dont see it being useful here tho

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

*teaching me something

Answer 26

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

Answer 27

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!!

Answer 28

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!!