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Differential Equation

  1. Dec 11, 2005 #1
    Hi all,

    Is the general solution of xy' + (1+x) y = e^(-x) sin 2x,
    y= (-cos2x)/(2xe^(x)) + c/(xe^(x))

    Thank you very much guys
  2. jcsd
  3. Dec 11, 2005 #2


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    Why don't you just plug it in so you can check for yourself whether it is correct?
  4. Dec 11, 2005 #3


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    Well the "c/(xe^(x))" part looks good.

  5. Dec 11, 2005 #4
    Because the other way arround is too long, and I'm not that good to garanty free-mistakes in the backward if I couldn't in the forward.
  6. Dec 11, 2005 #5
    I solved it again I got (-cos2x)/(2xe^(x)) that (1/2)(1/x)(cos2x/e^x)
  7. Dec 11, 2005 #6


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    If you won't do it "backward" then you'll never know if your "forward" is right. My advice: Just do it!
  8. Dec 12, 2005 #7


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    Well, first write it as:



    Then solve for the integrating factor:

    [tex]\sigma=Exp[\int (1+1/x)dx][/tex]



    multiplying both sides of the DE by this integrating factor results in:

    [tex]d\left[xe^x y]=Sin[2x]dx[/tex]

    Can you finish it now?
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