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Reducing the Bernoulli Equation

  1. May 29, 2012 #1


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    I'm practicing first-order linear differential equations, and have come across something I find interesting - being able to reduce nonlinear equations to linear equation with appropriate substitutions. I'll start with the well-known Bernoulli equation, and if there are other ways to do this technique please share!

    [itex]y' + P(x)y = Q(x)y^n[/itex]

    This equation is linear when [itex]n=0[/itex] and has seperable variables if [itex]n=1[/itex]. So, in the following development, and assuming that [itex]n≠0[/itex] and [itex]n≠1[/itex], we can multiply by [itex]y^{-n}[/itex] and [itex](1-n)[/itex] to obtain

    [itex]y^{-n}y' + P(x)y^{1-n}=Q(x)[/itex]



    which is a linear equation with the variable [itex]y^{1-n}[/itex], and if we let [itex]z=y^{1-n}[/itex] we then get


    Now, by multiplying by the integrating factor [itex]e^{∫P(x)dx}[/itex] we can convert the left side of the equation into the derivative of the product [itex]ye^{∫P(x)dx}[/itex], and we get the general solution of the Bernoulli equation!


    That's freakin' qute, right?
  2. jcsd
  3. May 29, 2012 #2
  4. May 29, 2012 #3


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    That it is! Wolfram demonstrates it a bit more eloquently, naturally.
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