solve: e^integral(-2/t)dt Any help is appreciated. Kind regards
\[\huge e^{\int -\frac 2t dt}\] do you know the value of \[\huge \int -\frac 2t dt\]
I think it is -2ln(t)
right
now apply the log rule.. \[\huge a \ln b \implies \ln b^a\]
so what will be the value of -2 ln t
-1 minute plse ...screen has gone blank
sure sure
busy powering up imac ...1 more minute plse
no rush :D
I'm assuming you want the integral of e^(2t) dt The integral is simply (1/2)*e^(2t) + c (c is a constant) This is because the integral of an exponential, e^(at) dt is: (1/a)e^(at) + c
for some reason openstudy wont open on imac...this could be a while ...sorry....let me log off on the windos pc...mebbe that the problem ;)
haha take your time. just tag me when you're ready
the answer is lnt^-2
right.. so.. \[\LARGE e^{\int -\frac 2t dt} \implies e^{-2 \ln t} \implies e^{\ln (t^{-2})}\]
okay
do you know how to simplify that?
yes, t^-2 = 1/t^2
right
does the ln cancel the e out
yup \[\huge e^{\ln a} = a\]
excellent - where do i find all these rules..do you know of a good resource
i have written a list of log rules before in this site...it also comes with a brief description and examples... http://openstudy.com/updates/4fa45f8fe4b029e9dc34e0b5
Wow...That's amazing....I didn't even know we had such excellent resources here on openstudy....Thank you very much ....Much appreciated
welcome ^_^ btw...you wouldnt be happening to be studying integrating factors in differential equations, would you?
yep...thats exactly why i posted this question..it is part of an Integrating Factor problem...How did you guess.?
because it didnt look like a normal integration problem =)) the only time i saw that kind of integration is from integrating factor
You are spot on....nice! Its not too hard now...we move on to 2nd order next semester. Is that difficult?
well...hmm...not really...you just have to remember solutions
okay...If you have any more tips on how to solve these IF problems that would be much appreciated. I struggle sometimes with the exp function...but with the help of the great people (like you) at openstudy my task has been made more easier... :)
hmm i dont know shortcuts to IF haha
okay...;) anyway ..thanks for your help .....i really aprreciate it...take care . :)
welcome ^_^
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