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Differential equiations

  1. Sep 5, 2009 #1
    the general solution for ye^x dy/dx = [e^-y ] + [ e^ -2x-y ]is:

    so i tried to see if this was a homogenous equation but it is not.

    next i tried to simplify and got:

    y e^y dy= (1+e^-2x) dx / e^x
     
  2. jcsd
  3. Sep 5, 2009 #2
    Could you please re-post this using the homework layout? Also, you are missing the equation that describes the general solution.

    Thanks
    Matt
     
  4. Sep 6, 2009 #3

    HallsofIvy

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    It is very difficult to understand what you mean. Do you mean that you are asked to find the general solution? And is "e^-2x- y" supposed to be "e^(-2x-y)" or "e^(-2x) - y"? Guessing at what you mean:
    Once you have
    [tex]y e^y dy= \frac{1+ e^{-2x}}{e^x} dx[/tex]
    just integrate both sides.

    Use integration by parts, letting u= y and [itex]dv= e^ydy[/itex] on the left.
    You can rewrite the right side as e-x+ e-3x which should be easy to integrate.
     
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