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Math Help - z=y' solution

  1. #1
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    z=y' solution

    xy''+2y'=x(y')^2
    z=y'
    xz'+2z=xz^2
    i cant solve it
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  2. #2
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    \displaystyle x\,\frac{dz}{dx} + 2z = x\,z^2

    \displaystyle \frac{dz}{dx} + 2x^{-1}z = z^2.


    This is a Bernoulli Equation, so make the substitution \displaystyle v = z^{1 - 2} = z^{-1}.

    Then \displaystyle z = v^{-1}, so

    \displaystyle \frac{dz}{dx} = \frac{d}{dx}(v^{-1})

    \displaystyle \frac{dz}{dx} = \frac{d}{dv}(v^{-1})\,\frac{dv}{dx}

    \displaystyle \frac{dz}{dx} = -v^{-2}\,\frac{dv}{dx}.


    Substituting into the DE \displaystyle \frac{dz}{dx} + 2x^{-1}z = z^2 gives

    \displaystyle -v^{-2}\,\frac{dv}{dx} + 2x^{-1}v^{-1} = (v^{-1})^2

    \displaystyle -v^{-2}\,\frac{dv}{dx} + 2x^{-1}v^{-1} = v^{-2}

    \displaystyle \frac{dv}{dx} - 2x^{-1}v = -1 .


    Now this is First Order Linear, so use the Integrating Factor method.

    The Integrating Factor is \displaystyle e^{\int{-2x^{-1}\,dx}} = e^{-2\ln{x}} = e^{\ln{(x^{-2})}} = x^{-2}.

    Multiplying through by the Integrating Factor gives

    \displaystyle x^{-2}\,\frac{dv}{dx} - 2x^{-3}v = -x^{-2}

    \displaystyle \frac{d}{dx}(x^{-2}v) = -x^{-2}

    \displaystyle x^{-2}v = \int{-x^{-2}\,dx}

    \displaystyle x^{-2}v = x^{-1} + C

    \displaystyle v = x + Cx^2.


    Since \displaystyle z = v^{-1}

    \displaystyle z = (x + Cx^2)^{-1} = \frac{1}{x + Cx^2} = \frac{1}{x(1 + Cx)}.


    And finally \displaystyle y = \int{z\,dx} = \int{\frac{1}{x(1+Cx)}\,dx}. You can do this using Partial Fractions.
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