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Who can find the solution to this ODE?

  1. Aug 14, 2009 #1
    Looking for the solution to the following ODE:
    [tex]
    \[
    \frac{dy}{dx} = \frac{x^2y^2 - y}{x}
    \]
    [/tex]
     
  2. jcsd
  3. Aug 14, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    rewrite it as

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


    then read about http://en.wikipedia.org/wiki/Bernoulli_differential_equation" [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Aug 14, 2009 #3
    I had not thought of turning it into a Bernoulli equation. Actually though I found out another, more elementary way of doing it using exact differentials:
    [tex]
    \[
    \frac{dy}{dx} = \frac{x^2y^2 - y}{x} \Rightarrow x dy = x^2y^2 dx - y dx
    \]
    [/tex]
    Divide through by [tex]y^2[/tex]
    [tex]
    \[
    x\frac{1}{y^2} dy = x^2 dx - \frac{1}{y} dx \Rightarrow x\frac{1}{y^2} dy - (- \frac{1}{y}) dx = x^2 dx
    \]
    \[
    \frac{x\frac{1}{y^2} dy - (- \frac{1}{y}) dx}{x^2} = dx
    \]
    [/tex]
    The LHS is the exact differential for the quotient [tex] \frac{-\frac{1}{y}}{x}[/tex]
    So we can integrate both sides to get
    [tex]
    \[
    -\frac{1}{xy} = x + c \Rightarrow y=-\frac{1}{x^2 + cx}
    \]
    [/tex]
     
  5. Aug 14, 2009 #4
    Nice Cody Palmer, very nice :approve:

    coomast
     
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