anonymous
  • anonymous
\[\int \frac{dx}{lnx}\]
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
How to start?
anonymous
  • anonymous
Do you know how to use integration by parts?
anonymous
  • anonymous
Integration by parts doesn't look nice..

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[\int \frac{dx}{lnx} = \frac{x}{lnx} - \int x d(\frac{1}{lnx})\]If that's what you meant...
amistre64
  • amistre64
u=lnx; x=e^u du=dx/x x du = dx \[\int \frac{1}{u}e^u~du\] maybe?
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=integrate+1%2Flnx you sure you got it right?
anonymous
  • anonymous
\[\int \frac{dx}{lnx}\]\[=\int \frac{x}{xlnx}dx\]\[=\int \frac{x}{lnx}d(lnx)\]\[=\int \frac{e^{lnx}}{lnx}d(lnx) \]\[=\int \frac{e^u}{u}du\]\[=\int \frac{1}{u}d(e^u)\].... 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
  • anonymous
Perhaps I'm in the wrong direction?!
amistre64
  • amistre64
it looks like any solution will have to run into a logramathic integral
anonymous
  • anonymous
What technique(s) is(are) required to solve this equation?
amistre64
  • amistre64
ones that i havent run into yet :/
amistre64
  • amistre64
you can try to develop a series poly for it, and see what that would bring you
anonymous
  • anonymous
:O I haven't learnt expanding the function!
amistre64
  • amistre64
what methods have you learnt?
anonymous
  • anonymous
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
anonymous
  • anonymous
*exact equation lol
amistre64
  • amistre64
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
anonymous
  • anonymous
Actually, how does Wronskian work. I just know it can determine the linear dependency.. Then, nothing :/
amistre64
  • amistre64
the wronskian is the short version of the undetermined coeffs
amistre64
  • amistre64
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
amistre64
  • amistre64
yeah, i dont see it being useful here tho
anonymous
  • anonymous
It...somehow... reminds me of Cramer's :/
amistre64
  • amistre64
it is, since the cramer is what you do at the end of the undetermined coeffs
anonymous
  • anonymous
Ahhhh.... Thanks for teaching something new though!!!
anonymous
  • anonymous
*teaching me something
amistre64
  • amistre64
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
anonymous
  • anonymous
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!!
anonymous
  • anonymous
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!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.