Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Reducing the Bernoulli Equation

  1. May 29, 2012 #1

    phion

    User Avatar
    Gold Member

    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]

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

    [itex]\frac{d}{dx}[/itex][itex][y^{1-n}]+(1-n)P(x)y^{1-n}=(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

    [itex]\frac{dz}{dx}+(1-n)P(x)z=(1-n)Q(x)[/itex].

    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!

    [itex]y^{(1-n)P(x)dx}=∫(1-n)Q(x)e^{∫(1-n)P(x)dx}dx+C[/itex]

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

    phion

    User Avatar
    Gold Member

    That it is! Wolfram demonstrates it a bit more eloquently, naturally.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Reducing the Bernoulli Equation
Loading...