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Question regarding Bernoulli Differential Equation

  1. Apr 8, 2013 #1
    In a video I was watching regarding how to solve these, the lecturer said that
    the form of a Bernoulli Differential Equation is y'+P(x)y=q(x)y^n
    where n>1

    This means that if n = 1, it wouldn't be a Bernoulli differential equation and would be a first order linear differential equation, but if n=1, y'+P(x)y=q(x)y doesn't take the form of y'+P(x)y=q(x), so how is this a first order linear differential equation?

    , here is the video in case anyone wants to check it out.
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Apr 8, 2013 #2
    y'+P(x)y=q(x)y
    y' + (P-q)y = 0

    so P is P-q, and q is 0.
     
  4. Apr 8, 2013 #3
    I don't quite understand what you did, if n=1

    y'(y^-1)+P(x)-q(x)=0
     
  5. Apr 8, 2013 #4
    standard form for a linear first order ode
    y' + f(x) y = g(x)

    a n=1 example of a Bernoulli like eqn
    is y' + P(x) y = q(x) y
    rewrite it as y' + (P(x) - q(x))y = 0.
    this is in standard form with f(x) = P(x)-q(x) and g(x) = 0.
     
  6. Apr 8, 2013 #5
    Ah, but my original question was whether this was a Bernoulli equation or a First Order Linear DFQ(if it is, why), b/c in the video, they stated n=1 as a First Order LinDFQ
     
  7. Apr 8, 2013 #6

    LCKurtz

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    Do you not understand that the above posts are explaining to you that it is a first order linear DE and why?
     
  8. Apr 8, 2013 #7
    Sorry, I don't. I'm not fluent enough to understand his proof without further explanation.

    edit:AH, I think I finally understand the proof.

    This means that the integrating factor would be e^(∫(fx)dx) correct?
     
    Last edited: Apr 8, 2013
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