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More differential equations

  1. Feb 16, 2006 #1


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    Gold Member

    I hate this stuff! :cry: :cry:

    The next problem I must face upon my path of enlightenment is the following:

    Solve the initial value problem:

    [tex] (2y - e^x )y' = ye^x ,y(0) = 1[/tex]

    I thought that it looked exact but I ended up with…

    [tex] \begin{array}{l}
    M = (2y - e^x )dy,N = (ye^x )dx \\
    \frac{{dM}}{{dx}} = - e^x ,\frac{{dN}}{{dy}} = e^x \\

    So it's not exact… how should I go about solving this? I really suck at these differentials by the way!
  2. jcsd
  3. Feb 16, 2006 #2


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    Staff Emeritus
    Science Advisor

    You have my permission to kick yourself in the behind!

    (2y- ex)y'= yex is the same as

    (2y- ex)dy= yexdx but in order to check if it is an exact differential, you must write it as Mdy+ Ndx= 0.
    That is: (2y-ex)dy- yexdx= 0
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