Thread: How to show an equation is homogeneous.

If the right side of the equation dy/dx = f(x,y) can be expressed as a function of the ratio y/x only, then the equation is said to be homogeneous.

(1) $\displaystyle \frac{dy}{dx} = \frac{x^2+xy+y^2}{x^2}$

$\displaystyle \frac{dy}{dx} = 1 + \frac{y}{x} + \frac{y^2}{x^2}$

That one was easy.

(2) $\displaystyle \frac{dy}{dx} = \frac{4y-3x}{2x-y}$

How do I show that this is homogeneous? I know there is a simple way, but I can't remember for the life of me. Thanks!

I had a crack at it and got this far:



There's probably a better way of doing it than this (mainly because this didn't get an answer).

hope this helps......somehow.

Originally Posted by shadow_2145

If the right side of the equation dy/dx = f(x,y) can be expressed as a function of the ratio y/x only, then the equation is said to be homogeneous.

(1) $\displaystyle \frac{dy}{dx} = \frac{x^2+xy+y^2}{x^2}$

$\displaystyle \frac{dy}{dx} = 1 + \frac{y}{x} + \frac{y^2}{x^2}$

That one was easy.

(2) $\displaystyle \frac{dy}{dx} = \frac{4y-3x}{2x-y}$

How do I show that this is homogeneous? I know there is a simple way, but I can't remember for the life of me. Thanks!

Please see the followings
$\displaystyle \frac{dy}{dx} = \frac{4y-3x}{2x-y}$
.....$\displaystyle =\frac{4y}{2x-y}-\frac{3x}{2x-y}$
.....$\displaystyle =\frac{4}{2\frac{x}{y}-1}-\frac{3}{2-\frac{y}{x}}$
.....$\displaystyle =\frac{4}{2\frac{1}{\frac{y}{x}}-1}-\frac{3}{2-\frac{y}{x}}$

For the second one, just divide by $\displaystyle x,$ top & bottom.