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Linear DE help.

  1. Nov 29, 2004 #1
    I've got a linear DE here,

    [tex] (x + 4y^2) dy + 2y dx =0 [/tex]

    I've tried to put it in the general form of a linear equation, and I would get,

    [tex] \frac {dy}{dx} + \frac {2y}{x+4y^2} = 0 [/tex]

    but I have problems isolating the x, so that I would get the [tex] P(x) [/tex] in the general form.
     
    Last edited: Nov 29, 2004
  2. jcsd
  3. Nov 29, 2004 #2

    AKG

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    This doesn't make sense. If "DE" stands for "differential equation" then no, you don't have a DE here.
     
  4. Nov 29, 2004 #3
    ^ sorry, its supposed to be equals to 0, i forgot to add that...
     
  5. Nov 29, 2004 #4

    Hurkyl

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    Think outside the box for a moment. There's a very easy way to turn this into an ODE.
     
  6. Nov 30, 2004 #5

    James R

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    Your D.E. is not linear, either.
     
  7. Nov 30, 2004 #6

    HallsofIvy

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    The way you have written it
    [tex] \frac {dy}{dx} + \frac {2y}{x+4y^2} = 0 [/tex]
    it is not linear.
    However, you can write
    [tex] (x + 4y^2) dy + 2y dx =0 [/tex]

    as
    [tex]\frac{dx}{dy}= -\frac{x+4y^2}{2y}= -\frac{1}{2y}x- 2y[/tex]
    which is a linear d.e. for x as a function of y.
     
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