From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: 1st O.D.E. w/ the Following 4 Qualities
Date: 26 Feb 1995 06:48:47 GMT

In article <3ip1tu$ln9@newsreader.wustl.edu>,
Elizabeth Ann Martin <emartin@artsci.wustl.edu> wrote:

>Does anyone know of a first order differential equation ...
	Although this really means  F(x, y, dy/dx) = 0  for some  F  of 3
	variables, all further adjectives assume the equation has the form
	dy/dx = f(x,y)  for some real-valued  f  defined on a subset of R^2.

>...that is linear, ...
	f(x,y) = g(x) y + h(x)  for some  g, h  of one real variable.

>...homogeneous, ...
	for all constants  c, f(cx, cy)=f(x,y) for all (x, y) in domain(f).
	With linearity this means [c g(cx) - g(x) ] y  + [h(cx)-h(x)] = 0;
	assuming  domain(f) is open, this requires  c g(cx)=g(x) and
	h(cx)=h(x). Allowing  c  to vary shows  h(x)=h(1) (constant) and
	g(x) = g(1) / x.  Thus f(x,y) = A y / x + B  for some constants  A,B.

>...and seperable? 
[That's   s-e-p-a-r-a-b-l-e.]
	f(x,y) = p(x)*q(y). This now requires  x p(x) q(y) =  A y + B x 
	for all  x  and  y. Holding  y  constant and varying x  shows that
	x p(x)  must be a linear function in  x, so  p(x) = m + b/x  for some
	constants  m  and  b.  Likewise holding  x  constant and varying y  
	shows  q(y) = m' y + b'. Then we need  (mx+b)(m'y+b') = A y + B x,
	which can only hold identically on an open set if  mm'=0 and  bb'=0.
	These require that either  p(x) = m and q(y) = b'  (so  f  is constant)
	or  p(x) = b/x  and q(y) = m'y  (so f(x,y) = const * (y/x).)

>How about dy/dx=C?
	It now must be either this or   dy/dx = C.(y/x).

You'll noticed I skipped "exact". Exactness isn't really a property of a
differential equation, but rather of a _presentation_ of a D.E.  Both
y dx + x dy = 0  and   (xy) dx + (x^2) dy = 0  are presentations of the
same  D.E  dy/dx = -y/x, but the first is exact and the second isn't.
That's the whole point of integrating factors -- they re-present a DE in
an exact form. As it turns out, both of the DE's above are easily written
in an exact form:  C dx - 1 dy = 0  and  [Cy x^(C-1)] dx - [x^C] dy = 0 
respectively.  (The form  C y dx - x dy = 0  is exact iff  C = -1).

dave

==============================================================================

Date: Fri, 6 Feb 1998 01:53:54 GMT
From: Robert Israel <israel@math.ubc.ca>
To: rusin@math.niu.edu

Note that "homogeneous" has (at least) two different meanings in differential
equations.  In the context of linear differential equations, "homogeneous"
usually means that there is no term without the dependent variable or one
of its derivatives.  So in general a first-order, linear, homogeneous DE would be 
of the form a(x) y' + b(x) y = 0.

... separable : implied by first-order linear homogeneous.

... exact : would mean that a'(x) = b(x).  

Cheers,
Robert