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Math Help - Where did I make a mistake? (1st order homogeneous ODE)

  1. #1
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    Where did I make a mistake? (1st order homogeneous ODE)

    I have a homogeneous ODE which I'm trying to solve:

    \frac{dy}{dx} = \frac{x^2 + y^2}{xy - x^2}


    Book answer:

    c(x + y)^2 = xe^{\frac{y}{x}}


    Here's what I'm doing:

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

    x \frac{dv}{dx} = \frac{v + 1}{v - 1}

    \int \frac{1}{x} dx = \int \frac{v - 1}{v + 1} dv

    ln(x) = \int 1 - \frac{2}{v + 1} dv

    ln(x) = v - 2ln(v + 1) + c

    x = \frac{e^v}{(v + 1)^2} c

    x = \frac{e^{\frac{y}{x}}}{(\frac{y}{x} + 1)^2} c

    cx = \frac{e^{\frac{y}{x}}}{(\frac{y + x}{x})^2}


    ...and this obviously looks nothing like the solution.

    Where did I make a mistake?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Where did I make a mistake? (1st order homogeneous ODE)

    Quote Originally Posted by Lancet View Post
    I have a homogeneous ODE which I'm trying to solve:

    \frac{dy}{dx} = \frac{x^2 + y^2}{xy - x^2}


    Book answer:

    c(x + y)^2 = xe^{\frac{y}{x}}


    Here's what I'm doing:

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

    x \frac{dv}{dx} = \frac{v + 1}{v - 1}

    \int \frac{1}{x} dx = \int \frac{v - 1}{v + 1} dv

    ln(x) = \int 1 - \frac{2}{v + 1} dv

    ln(x) = v - 2ln(v + 1) + c

    x = \frac{e^v}{(v + 1)^2} c

    x = \frac{e^{\frac{y}{x}}}{(\frac{y}{x} + 1)^2} c

    cx = \frac{e^{\frac{y}{x}}}{(\frac{y + x}{x})^2}


    ...and this obviously looks nothing like the solution.

    Where did I make a mistake?

    I'ts the same!


    cx = \frac{e^{\frac{y}{x}}}{(\frac{y + x}{x})^2}



    cx(\frac{y + x}{x})^2 = {e^{\frac{y}{x}}}





    cx(\frac{(y + x)^2}{x^2}) = {e^{\frac{y}{x}}}



    c(\frac{(y + x)^2}{x}) = {e^{\frac{y}{x}}}



    (\frac{(y + x)^2}{x}) = \frac{{e^{\frac{y}{x}}}}{c}



    (y + x)^2 = x\frac{{e^{\frac{y}{x}}}}{c}


    \frac{1}{c}=C



    (y + x)^2 = Cx{e^{\frac{y}{x}}}
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  3. #3
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    Re: Where did I make a mistake? (1st order homogeneous ODE)

    Quote Originally Posted by Also sprach Zarathustra View Post
    I'ts the same!
    That was a pleasant surprise!

    How do you get there from here, if you don't mind my asking?
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  4. #4
    Senior Member bugatti79's Avatar
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    Re: Where did I make a mistake? (1st order homogeneous ODE)

    Quote Originally Posted by Also sprach Zarathustra View Post
    I'ts the same!


    cx = \frac{e^{\frac{y}{x}}}{(\frac{y + x}{x})^2}



    cx(\frac{y + x}{x})^2 = {e^{\frac{y}{x}}}





    cx(\frac{(y + x)^2}{x^2}) = {e^{\frac{y}{x}}}



    c(\frac{(y + x)^2}{x}) = {e^{\frac{y}{x}}}



    (\frac{(y + x)^2}{x}) = \frac{{e^{\frac{y}{x}}}}{c}



    (y + x)^2 = x\frac{{e^{\frac{y}{x}}}}{c}


    \frac{1}{c}=C



    (y + x)^2 = Cx{e^{\frac{y}{x}}}
    In the 3rd last line of his post he has (y + x)^2 = x\frac{{e^{\frac{y}{x}}}}{c}

    which is the same as c(y + x)^2 = xe^{\frac{y}{x}} ie your books answer.
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  5. #5
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    Re: Where did I make a mistake? (1st order homogeneous ODE)

    I understand now! Thanks, guys!
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