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Another O.D.E

  1. Oct 17, 2005 #1
    for the following question:

    my problem:
    so (1/F)(dF/dx)=2
    so e^2x-(e^2x)(6e^y-2x)dy=0
    so (integration)(d^2x)dx=(1/2)e^2x +h(y)
    =>h(y)= -6e^(2x+y)+xye^2x+c
    so (1/2)e^2x-6e^(2x+y)+xye^2x+c-0

    however the correct answer should be

    does anybody know where my calculations went wrong?
  2. jcsd
  3. Oct 17, 2005 #2


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    Your correct answer hasn't got an x-term in it. Is that a typo ?
  4. Oct 18, 2005 #3
    no, it's not a typo~
  5. Oct 19, 2005 #4


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    Did a little research on this. Finally figured out what you were trying to do
    You have a differential equation and you were trying to force it to be an exact DE, yes ?

    I'm afraid you got the integrating factor wrong.
    You had,

    (1/F)dF/dx = 2,

    that should have been

    (1/F(y))dF(y)/dy = 2


    F(y) = e^(3y)

    Also, your "correct answer" is wrong,

    [tex]y = ce^{-2y} + 2e^y[/tex]

    There is no x-term in this eqn, so you have no relation between x and y, so you can't use this eqn to get a DE that involves an x, since x was never involved in the first place.

    The actual answer is,

    [tex]c = x.e^{2y} - 2e^{3y}[/tex]
    [tex]x = c.e^{-2y} + 2e^y[/tex]

    If you differentiate that, you'll end with the DE you started with.
    Last edited: Oct 19, 2005
  6. Oct 19, 2005 #5


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    HINT:Make [itex] e^{y(x)}=t(x) [/itex]

  7. Oct 19, 2005 #6
    opps... i corrected the integrating factor and it works! f=e^2y~
    why did you think of making [itex] e^{y(x)}=t(x) [/itex]?
  8. Oct 19, 2005 #7


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    I guess that's a substitution you get to know with experience ?

    It's faster/simpler than doing the exact differential thing.
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