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Math Help - Partial differentiation question

  1. #1
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    Partial differentiation question

    I have an equation of the form

    x\frac{{\delta q}}{{\delta y}} + y\frac{{\delta q}}{{\delta x}} = 0.

    How would I go about showing that q = f\left( {x/y} \right) is a solution? Can I say that \frac{{\delta q}}{{\delta y}} = \frac{{ - x}}{{{y^2}}}, and continue like that? I'm a bit confused with the fact that it's q = f\left( {x/y} \right) and not just q = x/y.

    Any help appreciated!
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  2. #2
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    Quote Originally Posted by ambience12 View Post
    I have an equation of the form

    x\frac{{\delta q}}{{\delta y}} + y\frac{{\delta q}}{{\delta x}} = 0.

    How would I go about showing that q = f\left( {x/y} \right) is a solution? Can I say that \frac{{\delta q}}{{\delta y}} = \frac{{ - x}}{{{y^2}}}, and continue like that? I'm a bit confused with the fact that it's q = f\left( {x/y} \right) and not just q = x/y.

    Any help appreciated!
    Here  f is an arbitrary function of its argument so

    \frac{\partial q}{\partial x} = f'\left( \frac{x}{y} \right) \frac{1}{y} (chain rule). Similarly \frac{\partial q}{\partial y} = f'\left( \frac{x}{y} \right) \left( - \frac{x}{y^2}\right).

    If you substitute you'll find that

    x\frac{{\partial q}}{{\partial y}} + y\frac{{\partial q}}{{\partial x}} = 0 identically.
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