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Math Help - Solve in two different ways

  1. #1
    MHF Contributor alexmahone's Avatar
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    Solve in two different ways

    Solve in two different ways \frac{dy}{dx}=xy^3-xy.

    -------------------------------------------------------------------------

    Separable:

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

    \frac{dy}{y^3-y}=xdx

    \frac{dy}{y(y+1)(y-1)}=xdx

    Let \frac{1}{y(y+1)(y-1)}=\frac{A}{y}+\frac{B}{y+1}+\frac{C}{y-1}

    1=A(0+1)(0-1)

    A=-1

    1=B(-1)(-1-1)

    B=\frac{1}{2}

    1=C(1+1)

    C=\frac{1}{2}

    \int \left[\frac{-1}{y}+\frac{1}{2(y+1)}+\frac{1}{2(y-1)}\right]dy=\int xdx

    -\ln y+\frac{1}{2}\ln (y+1)+\frac{1}{2}\ln (y-1)=\frac{x^2}{2}+C'

    -2\ln y+\ln (y+1)+\ln (y-1)=x^2+C, where C=2C'.

    \ln \frac{(y+1)(y-1)}{y^2}=x^2+C

    Bernoulli:

    \frac{dy}{dx}+xy=xy^3

    Let v=y^{-2}.

    y=v^{-1/2}

    \frac{dy}{dx}=-\frac{1}{2}v^{-3/2}\frac{dv}{dx}

    -\frac{1}{2}v^{-3/2}\frac{dv}{dx}+xv^{-1/2}=xv^{-3/2}

    Multiplying throughout by -2v^{3/2},

    \frac{dv}{dx}-2xv=-2x

    The integrating factor is e^{\int -2xdx}=e^{-x^2}.

    e^{-x^2}\frac{dv}{dx}-2xe^{-x^2}v=-2xe^{-x^2}

    \frac{d}{dx}(ve^{-x^2})=-2xe^{-x^2}

    ve^{-x^2}=e^{-x^2}+C

    y^{-2}e^{-x^2}=e^{-x^2}+C

    y^{-2}=Ce^{x^2}+1

    y=(Ce^{x^2}+1)^{-1/2}

    -------------------------------------------------------------------------

    Where have I gone wrong?
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  2. #2
    MHF Contributor
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    Re: Solve in two different ways

    Quote Originally Posted by alexmahone View Post
    Solve in two different ways \frac{dy}{dx}=xy^3-xy.

    -------------------------------------------------------------------------

    Separable:

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

    \frac{dy}{y^3-y}=xdx

    \frac{dy}{y(y+1)(y-1)}=xdx

    Let \frac{1}{y(y+1)(y-1)}=\frac{A}{y}+\frac{B}{y+1}+\frac{C}{y-1}

    1=A(0+1)(0-1)

    A=-1

    1=B(-1)(-1-1)

    B=\frac{1}{2}

    1=C(1+1)

    C=\frac{1}{2}

    \int \left[\frac{-1}{y}+\frac{1}{2(y+1)}+\frac{1}{2(y-1)}\right]dy=\int xdx

    -\ln y+\frac{1}{2}\ln (y+1)+\frac{1}{2}\ln (y-1)=\frac{x^2}{2}+C'

    -2\ln y+\ln (y+1)+\ln (y-1)=x^2+C, where C=2C'.

    \ln \frac{(y+1)(y-1)}{y^2}=x^2+C

    Bernoulli:

    \frac{dy}{dx}+xy=xy^3

    Let v=y^{-2}.

    y=v^{-1/2}

    \frac{dy}{dx}=-\frac{1}{2}v^{-3/2}\frac{dv}{dx}

    -\frac{1}{2}v^{-3/2}\frac{dv}{dx}+xv^{-1/2}=xv^{-3/2}

    Multiplying throughout by -2v^{3/2},

    \frac{dv}{dx}-2xv=-2x

    The integrating factor is e^{\int -2xdx}=e^{-x^2}.

    e^{-x^2}\frac{dv}{dx}-2xe^{-x^2}v=-2xe^{-x^2}

    \frac{d}{dx}(ve^{-x^2})=-2xe^{-x^2}

    ve^{-x^2}=e^{-x^2}+C

    y^{-2}e^{-x^2}=e^{-x^2}+C

    y^{-2}=Ce^{x^2}+1

    y=(Ce^{x^2}+1)^{-1/2}

    -------------------------------------------------------------------------

    Where have I gone wrong?
    What makes you think there's something wrong with either of them (except for leaving off the modulus signs in your logarithms)? If you try to write your separable equation as y in terms of x, you should get something that is correct up to an integration constant.
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Re: Solve in two different ways

    Quote Originally Posted by Prove It View Post
    What makes you think there's something wrong with either of them (except for leaving off the modulus signs in your logarithms)? If you try to write your separable equation as y in terms of x, you should get something that is correct up to an integration constant.
    \ln \left|\frac{(y+1)(y-1)}{y^2}\right|=x^2+C

    \frac{(y+1)(y-1)}{y^2}=\pm e^{(x^2+C)}=\pm e^Ce^{x^2}

    \frac{(y+1)(y-1)}{y^2}=C'e^{x^2}, where C'=\pm e^C

    y^2-1=C'e^{x^2}y^2

    y^2(1-C'e^{x^2})=1

    y^2=(1-C'e^{x^2})^{-1}

    y=(1-C'e^{x^2})^{-1/2}, where C' is the negative of the constant in the Bernoulli method.
    Last edited by alexmahone; September 26th 2011 at 01:01 AM.
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