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 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 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