a differential equation

liyi

Solve
$\frac{dy}{dx} = \frac{3x^2+2y^2-xy}{3x^2+2y^2+xy}$

**Posted:** Wed Apr 27, 2005 12:57 am

Tr

rewrite as
$\frac{dy}{dx} = \frac{(3\frac{x}{y}+2\frac{y}{x}-1)}{(3\frac{x}{y}+2\frac{y}{x}+1)}$
make the subtitution $y = vx$ ..change of variable
so $\frac{dy}{dx}=v+x\frac{dv}{dx}$

from then on it easy to seperate it and integrate to find v as a function of x and then multiply by x to find y the orginal function.

**Posted:** Wed Apr 27, 2005 12:25 pm

_________________
>**Tr**<

kunny

**Posted:** Wed Apr 27, 2005 6:27 pm

Dr Sonnhard Graubner

hello, with $y=xv$ i have got
$\frac{3+2v^2+v}{3+v^2-4v-2v^3}\,dv=\frac{dx}{x}$
Sonnhard.

**Posted:** Fri Oct 30, 2009 12:54 pm