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Math Help - Bernoulli's Equation - Fluid Flow Problem

  1. #1
    Member VitaX's Avatar
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    Bernoulli's Equation - Fluid Flow Problem

    y' = \epsilon y - \sigma y^3 ; \epsilon > 0 and \sigma > 0

    Show that if n does not equal 0 or 1, then the substitution v = y^{1 - n} reduces Bernoulli's equation to a linear equation.


    Bernoulli's Equation Standard Form y' + p(t)y = q(t)y^n


    y' - \epsilon y = -\sigma y^3 \rightarrow y'y^{-3} - \epsilon y^{-2} = -\sigma


    n = 3 ; v = y^{1-3} = y^{-2} \rightarrow dv = -2y^{-3} dy \rightarrow -\frac{1}{2}dv = y^{-3}dy


    -\frac{1}{2}v' - \epsilon v = -\sigma (Linear)


    v' + 2\epsilon v = 2\sigma


    \mu = e^{\int 2\epsilon} = e^{2\epsilon t}


    e^{2\epsilon t}[v' + 2\epsilon v] = 2\sigma e^{2\epsilon t}


    \frac{dv}{dt}[e^{2\epsilon t}v] = 2\sigma e^{2\epsilon t}


    \int dv[e^{2\epsilon t}v] = \int [2\sigma e^{2\epsilon t}]dt


    e^{2\epsilon t}v = 2\sigma \int e^{2\epsilon t}dt \rightarrow e^{2\epsilon t}v = 2\sigma \left[\frac{e^{2\epsilon t}}{2\epsilon}}\right]


    v = \frac{\sigma e^{2\epsilon t}}{\epsilon e^{2\epsilon t}} \rightarrow v = \frac{\sigma}{\epsilon}


    y^{-2} = \frac{\sigma}{\epsilon}

    y = \pm \sqrt{\frac{\epsilon}{\sigma}

    The books answer was
    y = \pm \sqrt{\frac{\epsilon}{\sigma + c\epsilon e^{-2\epsilon t}}}

    My answer's not even close, can someone walk me through this problem so I can see what I did wrong?
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  2. #2
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    Re: Bernoulli's Equation - Fluid Flow Problem

    Where's your constant of integration?
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  3. #3
    Member VitaX's Avatar
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    Re: Bernoulli's Equation - Fluid Flow Problem

    Can't believe forgetting the constant of integration had such an effect on the final answer. Well anyways, I got y = \pm \sqrt{\frac{\epsilon e^{2\epsion t}}{\sigma e^{2\epsilon t} + c\epsilon}}

    Is my answer equivalent to there's? It's kind of hard to see. y = \pm \sqrt{\frac{\epsilon}{\sigma + c\epsilon e^{-2\epsilon t}}}
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    Re: Bernoulli's Equation - Fluid Flow Problem

    Multiply top and bottom of the fraction inside the square root sign with e^{-2\epsilon t}. I think the exponential in your numerator might be off.
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  5. #5
    Member VitaX's Avatar
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    Re: Bernoulli's Equation - Fluid Flow Problem

    Oh forgot the epsilon in it.
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    Re: Bernoulli's Equation - Fluid Flow Problem

    So, you're all set now?
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