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Thread: Partial differentiation question

  1. #1
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    Partial differentiation question

    I have an equation of the form

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

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

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

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

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

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

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

    If you substitute you'll find that

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