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Math Help - Bernulli equations

  1. #1
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    Bernulli equations

    Please help me to solve next equations:
    1. ydx+(x-\frac{1}{2}x^{3}y)dy=0<br />

    <br /> <br />
2.x=(x^{2}-2y+1)y'<br /> <br />
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  2. #2
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    Quote Originally Posted by sinichko View Post
    Please help me to solve next equations:
    1. ydx+(x-\frac{1}{2}x^{3}y)dy=0<br />

    <br /> <br />
2.x=(x^{2}-2y+1)y'<br /> <br />
    y\,dx + \left(x - \frac{1}{2}\,x^3y\right)dy = 0

    \left(x - \frac{1}{2}\,x^3y\right)\frac{dy}{dx} + y = 0

    \left(\frac{2x - x^3y}{2}\right)\frac{dy}{dx} + y = 0

    \left(\frac{2x - x^3y}{2}\right)\frac{dy}{dx} = -y

    \frac{dy}{dx} = \frac{2y}{x^3y - 2x}

    \frac{dx}{dy} = \frac{x^3y - 2x}{2y}

    \frac{dx}{dy} = \frac{x^3y}{2y} - \frac{2x}{2y}

    \frac{dx}{dy} = \frac{1}{2}\,x^3 - \frac{1}{y}\,x

    \frac{dx}{dy} + \frac{1}{y}\,x = \frac{1}{2}\,x^3.


    Now this is a Bernoulli Equation, so make the substitution v = x^2, so x = v^{\frac{1}{2}}.

    Therefore \frac{dx}{dy} = \frac{dx}{dv}\,\frac{dv}{dy}

    \frac{dx}{dy} = \frac{1}{2}\,v^{-\frac{1}{2}}\,\frac{dv}{dy}.


    Substituting into the DE gives:

    \frac{dx}{dy} + \frac{1}{y}\,x = \frac{1}{2}\,x^3

    \frac{1}{2}\,v^{-\frac{1}{2}}\,\frac{dv}{dy} + \frac{1}{y}\,v^{\frac{1}{2}} = \frac{1}{2}\,v^{\frac{3}{2}}

    \frac{dv}{dy} + \frac{2}{y}\,v = v^2.


    This is now first order linear. So the integrating factor is e^{\frac{2}{y}\,dy} = e^{2\ln{y}} = e^{\ln{y^2}} = y^2.

    Multiplying through by the integrating factor:

    \frac{dv}{dy} + \frac{2}{y}\,v = v^2

    y^2\,\frac{dv}{dy} + 2y\,v = y^2v^2

    \frac{d}{dy}\left(v\,y^2\right) = y^2v^2

    v\,y^2 = \int{y^2v^2\,dy}

    v\,y^2 = \frac{v^2y^3}{3} - \int{\frac{2v\,y^3}{3}\,dy}

    v\,y^2 = \frac{v^2y^3}{3} - \frac{2}{3}\int{v\,y^3\,dy}

    v\,y^2 = \frac{v^2y^3}{3} - \frac{2}{3}\left(\frac{v\,y^4}{4} - \int{\frac{y^4}{4}\,dy}\right)

    v\,y^2 = \frac{v^2y^3}{3} - \frac{v\,y^4}{6} + \frac{1}{6}\int{y^4\,dy}

    v\,y^2 = \frac{v^2y^3}{3} - \frac{v\,y^4}{6} + \frac{y^5}{30} + C.


    And since v = x^2

    x^2y^2 = \frac{x^4y^3}{3} - \frac{x^2y^4}{6} + \frac{y^5}{30} + C

    x^2y^2 - \frac{x^4y^3}{3} + \frac{x^2y^4}{6} - \frac{y^5}{30} = C.


    Edit: Actually it looks like I've done something wrong... I'll keep checking...
    Last edited by Prove It; April 10th 2010 at 09:57 AM.
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