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Thread: function of y second order equation 2

  1. #1
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    function of y second order equation 2

    $\displaystyle 2(y')^2=y''(y-1)$
    $\displaystyle y(1)=2 y'(1)=-1$

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

    $\displaystyle 2\,\dfrac{y'}{y-1}=\dfrac{y''}{y'},$ 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

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


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

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

    Now integrating both sides gives

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

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

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

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

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

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

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

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

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


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

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

    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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