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Math Help - ODE Proof #2

  1. #1
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    ODE Proof #2

    Problem:
    If F is homogeneous of degree k in x and y, F can be written in the form:

    F=x^k g\left(\frac{y}{x}\right)

    Use that to prove the following equation for F:

    xF_x+yF_y=kF

    Solution:
    I found that:

    xF_x=kx^kg\left(\frac{y}{x}\right)-x^{k-1}y g_x\left(\frac{y}{x}\right)

    and

    yF_y=x^{k-1} y g_y\left(\frac{y}{x}\right)

    Adding them:

    xF_x+yF_y=kx^k g\left(\frac{y}{x}\right)-x^{k-1}y g_x\left(\frac{y}{x}\right)+x^{k-1} y g_y\left(\frac{y}{x}\right)

    Now, I stopped!

    any help?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Liverpool View Post
    Problem:
    If F is homogeneous of degree k in x and y, F can be written in the form:

    F=x^k g\left(\frac{y}{x}\right)

    Use that to prove the following equation for F:

    xF_x+yF_y=kF

    Solution:
    I found that:

    xF_x=kx^kg\left(\frac{y}{x}\right)-x^{k-1}y g_x\left(\frac{y}{x}\right)

    and

    yF_y=x^{k-1} y g_y\left(\frac{y}{x}\right)

    Adding them:

    xF_x+yF_y=kx^k g\left(\frac{y}{x}\right)-x^{k-1}y g_x\left(\frac{y}{x}\right)+x^{k-1} y g_y\left(\frac{y}{x}\right)

    Now, I stopped!

    any help?
    In order to finish this, you would have to assume that g_x(y/x)=g_y(y/x).

    So maybe you're supposed to prove that xF_x+yF_y=kF \iff g_x(y/x)=g_y(y/x)?
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  3. #3
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    Nope.
    The question is as i write it.
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  4. #4
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    Chris is right! Another way to look at it is since g(\cdot) is a function of a single argument then

    \dfrac{\partial}{\partial x} \;g \left(\dfrac{y}{x}\right) = - \dfrac{y}{x^2} g ' \left(\dfrac{y}{x}\right) and \dfrac{\partial}{\partial y} \;g \left(\dfrac{y}{x}\right) =  \dfrac{1}{x} g ' \left(\dfrac{y}{x}\right).
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  5. #5
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    Maybe there is something missing in the problem.
    I will ask my doctor.
    Thanks.
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