# 1st order homogeneous D.E.

• Jan 31st 2010, 04:48 PM
BCHurricane89
1st order homogeneous D.E.
$\displaystyle (x+y)\frac{dy}{dx}=x-y$
both homogeneous of degree 0, and I need to use the subs: y=vx dy=vdx+xdv, but I get stuck what exactly to do.

I got it down to this: $\displaystyle x(1-v)dx-x(1+v)vdx+xdv=0$ maybe im not seeing something, not exactly sure.
• Jan 31st 2010, 04:55 PM
whitepenguin
hi I think...
$\displaystyle v= \frac {y}{x}$

$\displaystyle \frac {dy}{dx} = v + \frac {dv}{dx}.x$

$\displaystyle \frac {dy}{dx} = \frac {1 -v}{1+v}$

$\displaystyle -> v+ \frac {dv}{dx}.x = \frac {1-v}{1+v}$

put v to the other side now we have seprable form
• Jan 31st 2010, 05:06 PM
BCHurricane89
not quite seeing where you are putting that into...sry
• Jan 31st 2010, 05:31 PM
whitepenguin
Hi,
$\displaystyle v+ \frac {dv}{dx}.x = \frac {1-v}{1+v}$

$\displaystyle \frac{dv}{dx}.x = \frac {1-v}{1+v} -v$

$\displaystyle \frac {dv}{dx}.x = \frac {1-v - v(1+v)}{1+v}$

$\displaystyle \frac {(1+v)dv}{1-v - v -v^2} = \frac {dx}{x}$