1st order homogeneous D.E.

**BCHurricane89** · January 31st 2010, 04:48 PM

$(x+y)\frac{dy}{dx}=x-y$
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.

**whitepenguin** · January 31st 2010, 04:55 PM

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

**BCHurricane89** · January 31st 2010, 05:06 PM

not quite seeing where you are putting that into...sry

**whitepenguin** · January 31st 2010, 05:31 PM

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

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

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

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

Thread: 1st order homogeneous D.E.

LinkBack

Thread Tools

Display

1st order homogeneous D.E.

Similar Math Help Forum Discussions

1st order homogeneous

Homogeneous First-order ODE.

Second order non-homogeneous ODE

1st order homogeneous ODE

3rd order ODE, help with homogeneous

Search Tags