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RolyPoly Group Title

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

  • one year ago
  • one year ago

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  1. RolyPoly Group Title
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    How to start?

    • one year ago
  2. Meepi Group Title
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    Do you know how to use integration by parts?

    • one year ago
  3. RolyPoly Group Title
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    Integration by parts doesn't look nice..

    • one year ago
  4. RolyPoly Group Title
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    \[\int \frac{dx}{lnx} = \frac{x}{lnx} - \int x d(\frac{1}{lnx})\]If that's what you meant...

    • one year ago
  5. amistre64 Group Title
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    u=lnx; x=e^u du=dx/x x du = dx \[\int \frac{1}{u}e^u~du\] maybe?

    • one year ago
  6. amistre64 Group Title
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    http://www.wolframalpha.com/input/?i=integrate+1%2Flnx you sure you got it right?

    • one year ago
  7. RolyPoly Group Title
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    \[\int \frac{dx}{lnx}\]\[=\int \frac{x}{xlnx}dx\]\[=\int \frac{x}{lnx}d(lnx)\]\[=\int \frac{e^{lnx}}{lnx}d(lnx) \]\[=\int \frac{e^u}{u}du\]\[=\int \frac{1}{u}d(e^u)\].... 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 :/

    • one year ago
  8. RolyPoly Group Title
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    Perhaps I'm in the wrong direction?!

    • one year ago
  9. amistre64 Group Title
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    it looks like any solution will have to run into a logramathic integral

    • one year ago
  10. RolyPoly Group Title
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    What technique(s) is(are) required to solve this equation?

    • one year ago
  11. amistre64 Group Title
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    ones that i havent run into yet :/

    • one year ago
  12. amistre64 Group Title
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    you can try to develop a series poly for it, and see what that would bring you

    • one year ago
  13. RolyPoly Group Title
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    :O I haven't learnt expanding the function!

    • one year ago
  14. amistre64 Group Title
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    what methods have you learnt?

    • one year ago
  15. RolyPoly Group Title
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    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

    • one year ago
  16. RolyPoly Group Title
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    *exact equation lol

    • one year ago
  17. amistre64 Group Title
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    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

    • one year ago
  18. RolyPoly Group Title
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    Actually, how does Wronskian work. I just know it can determine the linear dependency.. Then, nothing :/

    • one year ago
  19. amistre64 Group Title
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    the wronskian is the short version of the undetermined coeffs

    • one year ago
  20. amistre64 Group Title
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    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

    • one year ago
  21. amistre64 Group Title
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    yeah, i dont see it being useful here tho

    • one year ago
  22. RolyPoly Group Title
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    It...somehow... reminds me of Cramer's :/

    • one year ago
  23. amistre64 Group Title
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    it is, since the cramer is what you do at the end of the undetermined coeffs

    • one year ago
  24. RolyPoly Group Title
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    Ahhhh.... Thanks for teaching something new though!!!

    • one year ago
  25. RolyPoly Group Title
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    *teaching me something

    • one year ago
  26. amistre64 Group Title
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    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

    • one year ago
  27. RolyPoly Group Title
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    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!!

    • one year ago
  28. RolyPoly Group Title
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    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!!

    • one year ago
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