# 1st order homogeneous D.E.

• January 31st 2010, 05:48 PM
BCHurricane89
1st order homogeneous D.E.
$(x+y)\frac{dy}{dx}=x-y$
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: $x(1-v)dx-x(1+v)vdx+xdv=0$ maybe im not seeing something, not exactly sure.
• January 31st 2010, 05:55 PM
whitepenguin
hi I think...
$v= \frac {y}{x}$

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

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

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

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

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

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

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