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Solving y'=ysin(2x) 
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#1
Jan2207, 11:15 PM

P: 21

I found the integrating factor to be e^x for DE y'=ysin(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^x2cos(2x)e^x4*integral(sin(2x)(e^x) I am stuck. Help! Thanks 


#2
Jan2207, 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.



#3
Jan2307, 12:28 AM

P: 21

directly? I think i am not seeing something



#4
Jan2307, 12:33 AM

P: 21

Solving y'=ysin(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



#5
Jan2307, 12:35 AM

P: 21

OH ok i got it....wasnt seeing that lol...THanks



#6
Jan2307, 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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