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Solving y'=y-sin(2x)

  • Thread starter koolrizi
  • Start date
21
0
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
 

Answers and Replies

466
4
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.
 
21
0
directly? I think i am not seeing something
 
21
0
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
 
21
0
OH ok i got it....wasnt seeing that lol...THanks
 
13
0
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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