Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83 (MC840302); site boring.UUCP Path: utzoo!watmath!clyde!burl!ulysses!mhuxr!mhuxb!mhuxn!mhuxm!mhuxj!houxm!whuxlm!harpo!decvax!genrad!panda!talcott!harvard!seismo!mcvax!boring!lambert From: lambert@boring.UUCP Newsgroups: net.math Subject: Re: Solution to a weighty problem? Message-ID: <6314@boring.UUCP> Date: Thu, 7-Feb-85 23:22:24 EST Article-I.D.: boring.6314 Posted: Thu Feb 7 23:22:24 1985 Date-Received: Sun, 10-Feb-85 05:29:31 EST References: <462@decwrl.UUCP> Reply-To: lambert@boring.UUCP (Lambert Meertens) Organization: CWI, Amsterdam Lines: 52 Summary: Apparently-To: rnews@mcvax.LOCAL The problem is to integrate y"(t) = K/(y^2). Put v = dy/dt. Then y" = dv/dt = (dy/dt).(dv/dy) = v.dv/dy, so v.dv/dy = K/y^2. We can now separate the variables: v.dv = (K/y^2).dy This is easy to integrate: (1/2)v^2 = K.(1/H - 1/y). The H is a constant, to be determined from the initial conditions. (H = 1/{(K1)^2/(2K)+1/K2}.) Now dy/dt = v = sqrt(2K.(1/H-1/y). We can again separate the variables: dt = dy/sqrt(2K.(1/H-1/y)) = (y/sqrt(2K.(y/H-1))).dy. By putting z = sqrt(2K.(y/H-1)) we obtain a more manageable form: y = H.(z^2/(2K)+1), dy = (H.z/K).dz, so dt = (H^2/K).(z^2/(2K)+1).dz. This is again easy to integrate: t-t0 = (H^2/K).(z^3/(6K)+z). Here, t0 is a constant like H, determined by the initial conditions. Unfortunately, this gives an implicit solution. It can be made explicit by solving this equation of the third degree algebraically. For details on how to do this, see the recent article <410@nbs-amrf.UUCP>. -- Lambert Meertens ...!{seismo,philabs,decvax}!lambert@mcvax.UUCP CWI (Centre for Mathematics and Computer Science), Amsterdam