# Help understanding homog. eq.

1. Oct 27, 2007

### fireandwater

Hi. I'm having a problem understanding how to solve non-linear homogeneous equations. For example, (x$$^{}2$$+y$$^{}2$$)dx +xydy = 0 ; x=1, y=1

I understand that y=xv, v=y/x, and dy=xdv +vdx

To sub in,
x$$^{}2$$ + (v$$^{}2$$x$$^{}2$$)dx +x$$^{}2$$v(vdx+xdv) = 0

Here's where I get lost:
x$$^{}2$$[(1+v$$^{}2$$)dx +v$$^{}2$$dx +xvdv] = 0 =>

x$$^{}2$$(1+v$$^{}2$$)dx +v$$^{}2$$dx+xvdv = 0 =>

1+2v$$^{}2$$dx +xvdv = 0 =>

1+2v$$^{}2$$dx = -xvdv =>

$$\int-dx/x$$ = int(vdv/1+2v^2) =>

-ln x = 1/4 ln (1+2v$$^{}2$$) +C =>

ln x$$^{}4$$ + ln(1+2v$$^{}2$$) +4c = A

I think I'm just getting lost in all the algebraic "cleaning up", but I can't figure it out. Can someone pick it apart for me?

2. Oct 28, 2007

### arildno

$$x^{2}dx+v^{2}x^{2}dx+x^{2}v(vdx+xdv)=0$$

Clean this up as follows:
$$(x^{2}+v^{2}x^{2}+x^{2}v^{2})dx+x^{3}vdv=0$$
That is to say:
$$x^{2}(1+2v^{2})dx+x^{3}vdv=0$$
or, divided by x^{2}:
$$(1+2v^{2})dx+xvdv=0$$

See if you follow this so far!