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RolyPoly

  • 2 years ago

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

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  1. RolyPoly
    • 2 years ago
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    How to start?

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

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

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

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

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

  7. RolyPoly
    • 2 years ago
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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 :/

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

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

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

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

  12. amistre64
    • 2 years ago
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    you can try to develop a series poly for it, and see what that would bring you

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

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

  15. RolyPoly
    • 2 years ago
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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

  16. RolyPoly
    • 2 years ago
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    *exact equation lol

  17. amistre64
    • 2 years ago
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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

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

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

  20. amistre64
    • 2 years ago
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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

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

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

  23. amistre64
    • 2 years ago
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    it is, since the cramer is what you do at the end of the undetermined coeffs

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

  25. RolyPoly
    • 2 years ago
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    *teaching me something

  26. amistre64
    • 2 years ago
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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

  27. RolyPoly
    • 2 years ago
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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!!

  28. RolyPoly
    • 2 years ago
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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!!

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