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Math Help - Nonlinear ODE

  1. #1
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    Nonlinear ODE

    Being a physicist, I rarely encountered a nonlinear ODE. Accordingly, I have no idea how to solve the following equation:

    y(x) * y''(x) + y'(x)^2 = 0

    with y(0) = C, y'(x) = 0.

    It's fairly obvious that one of the solutions is y(x) = C. But, how to find the other solution? I don't even know enough about general ODE to know is there another solution being the equation is not linear?

    Thanks

    *** I now see I should've posted this under the Differential Equations. Sorry! ***
    Last edited by Jester; January 2nd 2011 at 08:34 AM. Reason: moved thread
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  2. #2
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    \displaystyle y\,\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0

    \displaystyle y\,\frac{d^2y}{dx^2} = -\left(\frac{dy}{dx}\right)^2

    \displaystyle \frac{d^2y}{dx^2} = -\frac{\left(\frac{dy}{dx}\right)^2}{y}

    \displaystyle \frac{\frac{d^2y}{dx^2}}{\frac{dy}{dx}} = -\frac{\frac{dy}{dx}}{y}


    Now let \displaystyle u = \frac{dy}{dx} so that \displaystyle \frac{du}{dx} = \frac{d^2y}{dx^2} so that

    \displaystyle \frac{1}{u}\,\frac{du}{dx} = -\frac{1}{y}\,\frac{dy}{dx}

    \displaystyle \int{\frac{1}{u}\,\frac{du}{dx}\,dx} = \int{-\frac{1}{y}\,\frac{dy}{dx}\,dx}

    \displaystyle \int{\frac{1}{u}\,du} = \int{-\frac{1}{y}\,dy}

    \displaystyle \ln{|u|} + C_1 = -\ln{|y|} + C_2

    \displaystyle \ln{|u|} + \ln{|y|} = C where \displaystyle C = C_2 - C_1

    \displaystyle \ln{|u\,y|} = C

    \displaystyle \ln{\left|y\,\frac{dy}{dx}\right|} = C

    \displaystyle \left|y\,\frac{dy}{dx}\right| = e^C

    \displaystyle y\,\frac{dy}{dx} = A where \displaystyle A = \pm e^C

    \displaystyle \int{y\,\frac{dy}{dx}\,dx} = \int{A\,dx}

    \displaystyle \int{y\,dy} = Ax + C_3

    \displaystyle \frac{y^2}{2} + C_4 = Ax + C_3

    \displaystyle \frac{y^2}{2} = Ax + B where \displaystyle B = C_3 - C_4

    \displaystyle y^2 = 2(Ax + B)

    \displaystyle y = \pm \sqrt{2(Ax + B)}.
    Last edited by Prove It; January 2nd 2011 at 08:16 AM.
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  3. #3
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    Might be a little easier to recognize that

     <br />
y y'' + y'^2 = \left(y y'\right)'<br />
.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Another way: the function

    F(y,y',y'')=yy''+(y')^2

    satisfies:

    F(\lambda y,\lambda y',\lambda y'')=\lambda^2 F(y,y',y'')

    that is, it is homogeneous. Using the substitution:

    y=e^{\int z\;dx},\quad (z=z(x))

    the new equation is a first order equation.

    Fernando Revilla
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