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Math Help - Bernoulli equation

  1. #1
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    Exclamation Bernoulli equation

    I am having trouble expanding this Bernoulli equation correctly to solve it. Any help to get me this far would be wonderful.
    x^2 dx/dy = 2y (e^1/x y^-1/2 -1)

    For those who think it looks confusing what inside the brackets is supposed to read is e to the power of 1/x, y to the power of -1/2 is all one term and all of that has minus 1.
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  2. #2
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    Quote Originally Posted by njr008 View Post
    I am having trouble expanding this Bernoulli equation correctly to solve it. Any help to get me this far would be wonderful.
    x^2 dx/dy = 2y (e^1/x y^-1/2 -1)

    For those who think it looks confusing what inside the brackets is supposed to read is e to the power of 1/x, y to the power of -1/2 is all one term and all of that has minus 1.
    So that it's easier to read, the DE is

    x^2\frac{dx}{dy} = 2y(e^{\frac{1}{x}}y^{-\frac{1}{2}} - 1).

    I'll have a go of it now...

    Edit: I'm going to assume that you've made a typo, because if it involves \frac{dx}{dy} it is near impossible to get it to the required form

    \frac{dx}{dy} + P(y)x = Q(y)x^n.

    So I'll fix the typo so that it should say \frac{dy}{dx} instead. Then it is easy to get to the required form

    \frac{dy}{dx} + P(x)y = Q(x)y^n.


    So, the original DE should be (I think...)

    x^2\frac{dy}{dx} = 2y(e^{\frac{1}{x}}y^{-\frac{1}{2}}-1)

    Some rearranging will give you

    \frac{dy}{dx} + 2x^{-2}y = 2y^{\frac{1}{2}}e^{\frac{1}{x}}x^{-2}.


    Here n = \frac{1}{2} so 1 - n = \frac{1}{2}.

    So make the substitution v = y^{\frac{1}{2}} \implies y = v^2


    Now find \frac{dy}{dx}. Notice that \frac{dy}{dx} = \frac{dy}{dv}\frac{dv}{dx}.

    So \frac{dy}{dx} = 2v\frac{dv}{dx}.


    So, when substituting everything into the DE, it becomes

    2v\frac{dv}{dx} + 2x^{-2}v^2 = 2ve^{\frac{1}{x}}x^{-2}

    \frac{dv}{dx} + x^{-2}v = e^{\frac{1}{x}}x^{-2}.


    This is first order linear, so can be solved using an integrating factor.

    I = e^{\int{x^{-2}\,dx}} = e^{-\frac{1}{x}}


    e^{-\frac{1}{x}}\frac{dv}{dx} + e^{-\frac{1}{x}}x^{-2}v = e^{-\frac{1}{x}}e^{\frac{1}{x}}x^{-2}

    \frac{d}{dx}\left(e^{-\frac{1}{x}}v\right) = x^{-2}

    e^{-\frac{1}{x}}v = \int{x^{-2}\,dx}

    e^{-\frac{1}{x}}v = -\frac{1}{x} + C

    v = -e^{\frac{1}{x}}\frac{1}{x} + Ce^{\frac{1}{x}}

    y^{\frac{1}{2}} = -e^{\frac{1}{x}}\frac{1}{x} + Ce^{\frac{1}{x}}

    y = \left(-e^{\frac{1}{x}}\frac{1}{x} + Ce^{\frac{1}{x}}\right)^2.


    Phew...
    Last edited by Prove It; April 18th 2009 at 12:17 AM.
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  3. #3
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    Please see edit...
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  4. #4
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    Quote Originally Posted by Prove It View Post
    Please see edit...
    i am also agree with ur point...
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