1. ## 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!

2. Originally Posted by ambience12
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.