## RolyPoly Group Title $\int \frac{dx}{lnx}$ one year ago one year ago

1. RolyPoly Group Title

How to start?

2. Meepi Group Title

Do you know how to use integration by parts?

3. RolyPoly Group Title

Integration by parts doesn't look nice..

4. RolyPoly Group Title

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

5. amistre64 Group Title

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

6. amistre64 Group Title

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

7. RolyPoly Group Title

$\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 :/

8. RolyPoly Group Title

Perhaps I'm in the wrong direction?!

9. amistre64 Group Title

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

10. RolyPoly Group Title

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

11. amistre64 Group Title

ones that i havent run into yet :/

12. amistre64 Group Title

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

13. RolyPoly Group Title

:O I haven't learnt expanding the function!

14. amistre64 Group Title

what methods have you learnt?

15. RolyPoly Group Title

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

16. RolyPoly Group Title

*exact equation lol

17. amistre64 Group Title

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

18. RolyPoly Group Title

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

19. amistre64 Group Title

the wronskian is the short version of the undetermined coeffs

20. amistre64 Group Title

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

21. amistre64 Group Title

yeah, i dont see it being useful here tho

22. RolyPoly Group Title

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

23. amistre64 Group Title

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

24. RolyPoly Group Title

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

25. RolyPoly Group Title

*teaching me something

26. amistre64 Group Title

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

27. RolyPoly Group Title

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!!

28. RolyPoly Group Title

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!!