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RolyPoly Group Title

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

  • 2 years ago
  • 2 years ago

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  1. RolyPoly Group Title
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    How to start?

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

    • 2 years ago
  3. RolyPoly Group Title
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    Integration by parts doesn't look nice..

    • 2 years 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...

    • 2 years 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?

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

    • 2 years 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 :/

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

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

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

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

    • 2 years 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

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

    • 2 years ago
  14. amistre64 Group Title
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    what methods have you learnt?

    • 2 years 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

    • 2 years ago
  16. RolyPoly Group Title
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    *exact equation lol

    • 2 years 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

    • 2 years 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 :/

    • 2 years ago
  19. amistre64 Group Title
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    the wronskian is the short version of the undetermined coeffs

    • 2 years 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

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

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

    • 2 years 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

    • 2 years ago
  24. RolyPoly Group Title
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    Ahhhh.... Thanks for teaching something new though!!!

    • 2 years ago
  25. RolyPoly Group Title
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    *teaching me something

    • 2 years 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

    • 2 years 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!!

    • 2 years 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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