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Solving y'=y-sin(2x)

by koolrizi
Tags: solving, yysin2x
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koolrizi
#1
Jan22-07, 11:15 PM
P: 21
I found the integrating factor to be e^x for DE y'=y-sin(2x)
Now i am stuck at integral of -sin(2x)e^x
Can you help me with it? I tried using integration by parts and i get to following
integral(sin(2x)(e^x)=sin(2x)e^x-2cos(2x)e^x-4*integral(sin(2x)(e^x)

I am stuck. Help! Thanks
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Manchot
#2
Jan22-07, 11:33 PM
P: 728
Notice that you have the expression "integral(sin(2x)(e^x))" on both sides of your equation. You can therefore solve for it directly, and be done.
koolrizi
#3
Jan23-07, 12:28 AM
P: 21
directly? I think i am not seeing something

koolrizi
#4
Jan23-07, 12:33 AM
P: 21
Solving y'=y-sin(2x)

The answer to this DE is 1/5(sin(2x)+2cos(2x) and i am not seeing how to get to this from what i have
koolrizi
#5
Jan23-07, 12:35 AM
P: 21
OH ok i got it....wasnt seeing that lol...THanks
joob
#6
Jan23-07, 04:21 AM
P: 13
also careful with your results. looks like you missed a couple small details. your integrating factor is actually $e^{-x}$ and you should have the boundaries of the LHS of your equation(unless they're supposed to be =0)


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