Partial differentiation question

**ambience12** · April 25th 2009, 02:14 PM

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!

**Danny** · April 25th 2009, 02:23 PM

Originally Posted by ambience12

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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