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anonymous
 3 years ago
\[\int \frac{dx}{lnx}\]
anonymous
 3 years ago
\[\int \frac{dx}{lnx}\]

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you know how to use integration by parts?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Integration by parts doesn't look nice..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int \frac{dx}{lnx} = \frac{x}{lnx}  \int x d(\frac{1}{lnx})\]If that's what you meant...

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2u=lnx; x=e^u du=dx/x x du = dx \[\int \frac{1}{u}e^u~du\] maybe?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2http://www.wolframalpha.com/input/?i=integrate+1%2Flnx you sure you got it right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int \frac{dx}{lnx}\]\[=\int \frac{x}{xlnx}dx\]\[=\int \frac{x}{lnx}d(lnx)\]\[=\int \frac{e^{lnx}}{lnx}d(lnx) \]\[=\int \frac{e^u}{u}du\]\[=\int \frac{1}{u}d(e^u)\].... 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 :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Perhaps I'm in the wrong direction?!

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2it looks like any solution will have to run into a logramathic integral

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What technique(s) is(are) required to solve this equation?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2ones that i havent run into yet :/

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2you can try to develop a series poly for it, and see what that would bring you

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0:O I haven't learnt expanding the function!

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2what methods have you learnt?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0separable, 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/1803differentialequationsspring2010/readings/notes_exe/MIT18_03S10_1ex.pdf 1A3a

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2y^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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Actually, how does Wronskian work. I just know it can determine the linear dependency.. Then, nothing :/

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2the wronskian is the short version of the undetermined coeffs

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2Wx 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2yeah, i dont see it being useful here tho

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It...somehow... reminds me of Cramer's :/

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2it is, since the cramer is what you do at the end of the undetermined coeffs

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ahhhh.... Thanks for teaching something new though!!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0*teaching me something

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2yep, 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Never mind, perhaps i should ask my teacher after the lesson today. I have to go for lesson now. Once again, thanks for your help!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry 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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