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Homework Help: First order differential equation

  1. Jan 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Solve the differential equation:

    2. Relevant equations

    1+(x-x^2*e^(2y))(dy/dx) = 0

    3. The attempt at a solution

    No idea how to approach this.
     
  2. jcsd
  3. Jan 13, 2010 #2

    HallsofIvy

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    It looks pretty clear to me. Basic algebra gives
    [tex]x(1- e^{2y})dy/dx= -1[/tex]
    and then separate variables as
    [tex](1- e^{2y})dy= -\frac{1}{x}dx[/tex]

    Now integrate.
     
  4. Jan 13, 2010 #3
    Hmm, there's an x^2 in the equation, I think you missed that bit.
     
  5. Jan 13, 2010 #4

    tiny-tim

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    Hi Romaha_1! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Does putting z = e2y help? :smile:
     
  6. Jan 13, 2010 #5
    No it doesn't, does it?
     
  7. Jan 13, 2010 #6
    Yes, that's the reason I don't know how to do it. I don't think it is a mistake in the problem, since there is a sign "(very tricky)" after it on the problem set.
     
  8. Jan 13, 2010 #7

    tiny-tim

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    Doesn't it? :smile:
     
  9. Jan 13, 2010 #8
    The question isn't necessarily wrong, I'm not too good at this stuff at all. It just seemed to me that HallsofIvy was attempting the separable ODE approach, but missed the x^2.

    I think it should be:

    (e^2y)*dy = ((-1-x)/(-x^2))*dx

    (e[tex]^{2y}[/tex])*dy = ([tex]\frac{1-x}{-x^{2}}[/tex])*dx

    after the separation.

    Now you must integrate it. I'm not too sure if I'm right, so be careful.
     
    Last edited: Jan 13, 2010
  10. Jan 13, 2010 #9
    Yeah, I understand what you meant and agree with you that the separable approach does not work; it was my assumption that the question was wrong.
     
  11. Jan 13, 2010 #10
    I never said the Separable ODE approach doesn't work.
     
  12. Jan 13, 2010 #11
    Sorry, for some reason I could not see the end of your previous post. But still, I do not understnad how this works because the whole expression (x-x2*e2y) is multiplied by dy/dx, not just x2*e2y.
     
  13. Jan 13, 2010 #12
    Hi tiny-tim, thank you for the idea!

    But I could solve it only when letting z = x2*e2y; I still do not see how this is possible with z = e2y.
     
  14. Jan 14, 2010 #13

    tiny-tim

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    ah … on this forum, we don't give you the full answer …

    so, since you had no idea, I sort-of pushed you half-way (-1 = (x - x2z)(1/2z)dz/dx), and left you to finish it. :wink:

    having said that, I don't see how z = x2*e2y does it … perhaps i'm misreading the question? :redface:
     
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