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Bernouli Equation

  1. Aug 6, 2009 #1
    Recently I've been going back in my differential equation book to review some differential equation solving skills, in particular bernouli, ricatti's, and clairaut's equations; simple things enough. However when doing the exercises I have kept running into a "problem" with one question.

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

    Now I get it into the form to use the bernouli equation:

    Divide by x

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

    Multiply by y2

    [tex]y^2 \frac{dy}{dx} + \frac{y^3}{x} = \frac{1}{x}[/tex]

    So [tex]w=y^3[/tex]

    [tex]\frac{dw}{dx}=3y^2\frac{dy}{dx}[/tex]

    So now the equation is [tex]\frac{dw}{dx} + \frac{3w}{x} = \frac{3}{x}[/tex]

    So the integrating factor is [tex]e^\int^\frac{3}{x}^d^x=x^3[/tex]

    So you get [tex]\frac{d(wx^3)}{dx}=3x^2[/tex]

    w=1, so [tex]y^3=w=1=y[/tex]

    Which one can see works easily when plugging into the original formula. However the book gives a different answer of [tex]y^3=1+cx^-^3[/tex] which I don't understand how one could get with bernouli (and my calculator gives me the same answer with deSolve)... any help?
     
  2. jcsd
  3. Aug 6, 2009 #2

    rock.freak667

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    Homework Helper

    Now if
    [tex]\frac{d(wx^3)}{dx}=3x^2[/tex]

    then [itex]wx^3= ?[/itex]

    When you get that, sub back w=y3 and then divide by x3
     
  4. Aug 7, 2009 #3
    ahhhh!!! I forgot the constant of integration :( how silly of me....
     
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