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Thread: 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:

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


    Book answer:

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


    Here's what I'm doing:

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

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

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

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

    $\displaystyle ln(x) = v - 2ln(v + 1) + c$

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

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

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


    ...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:

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


    Book answer:

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


    Here's what I'm doing:

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

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

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

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

    $\displaystyle ln(x) = v - 2ln(v + 1) + c$

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

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

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


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

    Where did I make a mistake?

    I'ts the same!


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



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





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



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



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



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


    \frac{1}{c}=C



    $\displaystyle (y + x)^2 = Cx{e^{\frac{y}{x}}}$
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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!


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



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





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



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



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



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


    \frac{1}{c}=C



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

    which is the same as $\displaystyle c(y + x)^2 = xe^{\frac{y}{x}}$ 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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