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Solving y'=y-sin(2x)

  1. Jan 22, 2007 #1
    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
     
  2. jcsd
  3. Jan 22, 2007 #2
    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.
     
  4. Jan 23, 2007 #3
    directly? I think i am not seeing something
     
  5. Jan 23, 2007 #4
    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
     
  6. Jan 23, 2007 #5
    OH ok i got it....wasnt seeing that lol...THanks
     
  7. Jan 23, 2007 #6
    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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