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Sep 25 2016 12:17pm


integrate u'1 with respect to x

is this even possible..
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Sep 25 2016 12:58pm
sooo I tried it on Matlab
Code

syms x
int((exp(3*x)*(x^2)*(x+1)^2 - 3*(x^2)*exp(2*x)*cos(x))/(exp(3*x)*(x^2+x+1)),x)


and heres the output...

Code
ans =

-int((exp(-3*x)*(3*x^2*exp(2*x)*cos(x) - x^2*exp(3*x)*(x + 1)^2))/(x^2 + x + 1), x)


why did it give an integral in the output??
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Sep 25 2016 01:27pm
Define the function as an anonymous function
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Sep 25 2016 04:57pm
Quote (eLeMeNt477 @ Sep 25 2016 02:27pm)
Define the function as an anonymous function


how
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Sep 25 2016 05:53pm
Quote (FamilyGuyViewer @ Sep 25 2016 06:57pm)
how


google it
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Sep 25 2016 06:02pm
pm xyzamp
he can Integrate This Disgusting > Function np
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Sep 25 2016 06:39pm
Quote (eLeMeNt477 @ Sep 25 2016 06:53pm)
google it


tried didnt work
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Sep 25 2016 06:42pm
Quote (FamilyGuyViewer @ Sep 25 2016 08:39pm)
tried didnt work


https://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html
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Oct 7 2016 05:20am
Would you mind posting the entire assignment?

Are you solving this ODE with the method of undetermined coefficients?

Oh my gosh sorry for bumping a 2 week old post zz.

This post was edited by maruthenoob on Oct 7 2016 05:27am
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Oct 7 2016 10:36am
hmmm
it can be broken into a number of terms including products and quotients
breaking it up into easier to deal with terms might help, or maybe it wont
don't feel like looking into this because even if it isn't hard it's gonna take a while :p
the e^3x would totally disappear in one term, and stay as an e^x in the denominator in another
that would simplify things a bit. other simplifications with the polynomial are possible as well

This post was edited by ringo794 on Oct 7 2016 10:38am
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