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Nonlinear first order ODE

  1. Jan 22, 2013 #1
    1. The problem statement, all variables and given/known data

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

    3. The attempt at a solution

    Put it into derivative form.

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

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

    [tex] \frac{dy}{dx} + \frac{-1}{x}y = \frac{1}{x^2}y^2 [/tex]

    I recognized this as a Bernoulli equation where n = 2.

    [tex] \frac{dy}{dx}\frac{1}{y^2} + \frac{-1}{x}\frac{1}{y} = \frac{1}{x^2} [/tex]

    Plan to make a substitution of [itex] v = y^{-1} [/itex] and [itex] \frac{dv}{dx} = -y^{-2} \frac{dy}{dx} [/itex]

    [tex]-\frac{dv}{dx} + \frac{-1}{x}v = \frac{1}{x^2}[/tex]

    [tex]\frac{dv}{dx} + \frac{1}{x}v = \frac{-1}{x^2}[/tex]

    [tex] u(x) = e^{\int{\frac{1}{x}}dx} [/tex]

    [tex] u(x) = x [/tex]

    [tex] \int{x\frac{dv}{dx} + v} = \int{\frac{-1}{x}} [/tex]

    LHS is product of product rule.

    [tex] xv = -lnx + c [/tex]

    Since v = y^(-1)

    [tex] \frac{x}{y} = ln{\frac{c}{x}} [/tex]

    You could arrange the above equation into various forms, but I see no way to arrange it into the form in the answer key:

    [tex] x + yln{|x|} = cy [/tex]

    I've checked again and again and can't see where I'm going wrong with this...
     
  2. jcsd
  3. Jan 22, 2013 #2
    Recall what the logarithm of a ratio is.
     
  4. Jan 22, 2013 #3

    HallsofIvy

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    Strictly speaking this should be
    [tex] \frac{x}{y} = ln{\left|\frac{c}{x}\right|} [/tex]
    and, of course, [itex]ln|c/x|= ln|c|- ln|x|[/itex]

     
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