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Bernoulli's Equation (ODEs)

  1. Jan 21, 2013 #1

    Wow, just realized I skipped a crucial step. Forgot to isolate y' while I was working out the problem and now I see you can't even isolate y' without going back to the original equation. Sorry, disregard the thread please.

    I'm taking an introductory ODE course (I've only taken up to Calculus 2) and so far throughout the class, the textbook has been following a general strategy of throwing magical solutions out of thin air at me for certain forms an ordinary differential equation may take.

    Throughout the sections of the book, I've had to take the time and look at outside sources for clarification behind the theory and I'm unsure of what to think in this one.

    My textbook says that for an ODE of the form:

    [tex] \frac{dy}{dx} + yP(x) = Q(x)y^n [/tex]

    If I divide by y^n,

    [tex] y^{-n}\frac{dy}{dx} + P(x)y^{1-n} = Q(x) [/tex]

    The simple substitution v = y^(1-n) will simplify the equation into a linear first order ODE which may be solved by a method I already know of.

    When I read this, I figured that since a substitution seems to me like a visual aid more than anything, the key to solving the initial equation is really just dividing by y^n; that is, the ODE can be solved with an integrating factor without a need for substitution. At least that was what I thought.

    So, I tried it out:

    [tex] \frac{dy}{dx} + yp(x) = q(x)y^n [/tex]

    [tex] y^{-n}\frac{dy}{dx} + p(x)y^{1-n} = q(x) [/tex]

    [tex] u(x) = e^{\int{p(x)dx}} = e^{P(x)}[/tex]

    P'(x) = p(x)

    Multiplying across by u(x):

    [tex] y^{-n}\frac{dy}{dx}e^{P(x)} + p(x)y^{1-n}e^{P(x)} = e^{P(x)}q(x) [/tex]

    The left hand side however, is not the product of the product rule.

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

    The expression differs by the (1-n) factor and I'm stumped. As far as I can tell, there isn't a way to solve an ODE of that form through algebra and the integrating factor, but I can't really make myself see the substitution as anything more than a visual aid, yet the substitution combined with the integrating factor does work.

    What's going wrong with all of this?
    Last edited: Jan 21, 2013
  2. jcsd
  3. Jan 21, 2013 #2
    Closed by request
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