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Bernoulli yay!

  1. Jan 27, 2007 #1

    Here it goes...

    [tex]y' + \frac{y}{x} = 3x^2y^2[/tex]

    This is a Bernoulli equation with [tex]P = \frac{1}{2}[/tex], [tex]Q = 3x^2[/tex], and [tex]n = 2[/tex]. We first divide through by [tex]y^2[/tex], obtaining...

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

    We substitute [tex]z = y^-^1[/tex] and

    [tex]\frac{dz}{dx} = -y^-^2\frac{dy}{dx} \Longrightarrow \frac{dy}{dx} = -y^2\frac{dz}{dx}[/tex]

    Substituting... [tex]-\frac{dz}{dx} + \frac{z}{x} = 3x^2 \Longrightarrow \frac{dz}{x} - \frac{z}{x} = -3x^2[/tex]

    I'll stop there... two things I don't get... how does [tex]P = \frac{1}{2}[/tex] and how does that substitution work...

    If you can help, I would really appreciate it.
  2. jcsd
  3. Jan 28, 2007 #2


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    Yes, for that sentence to make any sense, you would have to know what a "Bernoulli equation" was, wouldn't you? Not remembering exactly and not having a differential equations text book immediately handy, I "googled" on "Bernoulli differential equation" and get (from Wikipedia)
    "an equation of the form y'+ P(x)y= Q(x)yn for n not equal to 1"

    There's obviously an error- either in your source or in your copying. That is a Bernoulli equation with P(x)= 1/x, not 1/2.

    Dividing a Bernoulli equation by yn and then making the substitution z= y1-n always reduces the equation to a linear equation.

    As for "how the substitution works" I don't know what more to say. It is shown pretty clearly there. What are you not understanding? (Other than that "P(x)= 1/2" should be P(x)= 1/x".)

    Do you see how they go from z= y-1 to dz/dx= -y-2dy/dx? That's straight forward "chain rule". Then, of course, y-2dy/dx in the differential equation becomes -dz/dx. y-1/x is simply z/x. The differential equation y-2dy/dx+ y-1/x= 3x2 becomes -dz/dx+ (1/x)z= 3x2.
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