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Math Help - function of y second order equation 2

  1. #1
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    function of y second order equation 2

    2(y')^2=y''(y-1)
    y(1)=2 y'(1)=-1

    again i tried
    y'=z
    z'y=z^{2}
    \frac{{dz}}{dx}=\frac{dz}{dy}\frac{dy}{dx}
    zz'=dz/dx
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  2. #2
    A Plied Mathematician
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    Try the same trick as on the other post:

    2\,\dfrac{y'}{y-1}=\dfrac{y''}{y'}, and integrate.
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  3. #3
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    how ?
    we dont have a single dy /dx member here
    we have many differentiated members on both side
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  4. #4
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    You have

    \displaystyle \frac{2}{y-1}\,\frac{dy}{dx} = \frac{1}{\frac{dy}{dx}}\,\frac{d^2y}{dx^2}.


    On the RHS make the substitution \displaystyle u = \frac{dy}{dx} so that \displaystyle \frac{du}{dx} = \frac{d^2y}{dx^2} so that your DE becomes

    \displaystyle \frac{2}{y-1}\,\frac{dy}{dx} = \frac{1}{u}\,\frac{du}{dx}.

    Now integrating both sides gives

    \displaystyle \int{\frac{2}{y-1}\,\frac{dy}{dx}\,dx} = \int{\frac{1}{u}\,\frac{du}{dx}\,dx}

    \displaystyle \int{\frac{2}{y-1}\,dy} = \int{\frac{1}{u}\,du}

    \displaystyle 2\ln{|y-1|} + C_1 = \ln{|u|} + C_2

    \displaystyle \ln{|y-1|^2} + C = \ln{|u|}, where \displaystyle C = C_1 - C_2

    \displaystyle C = \ln{|u|} - \ln{|(y-1)^2|}

    \displaystyle C = \ln{\left|\frac{u}{(y-1)^2}\right|}

    \displaystyle C = \ln{\left|\frac{\frac{dy}{dx}}{(y-1)^2}\right|}

    \displaystyle \pm e^C = \frac{dy}{dx}\,\frac{1}{(y-1)^2}

    \displaystyle A = \frac{dy}{dx}\,\frac{1}{(y-1)^2}, where \displaystyle A = \pm e^C.


    You can integrate both sides with respect to \displaystyle x now to find \displaystyle y. You will need to use partial fractions.
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  5. #5
    A Plied Mathematician
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    From my previous post:

    2\ln|y-1|=\ln|y'|+C_{1}.

    On both sides, you see, the derivative of the denominator appears in the numerator. Whenever that happens, you can do a u substitution, and you'll get du/u, which integrates to ln|u|. See how that works?
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