Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site ut-ngp.UUCP Path: utzoo!watmath!clyde!cbosgd!ihnp4!zehntel!dual!mordor!ut-sally!ut-ngp!lindley From: lindley@ut-ngp.UUCP (John L. Templer) Newsgroups: net.math Subject: Re: Solution to a weighty problem? Message-ID: <1271@ut-ngp.UUCP> Date: Sat, 2-Feb-85 20:42:05 EST Article-I.D.: ut-ngp.1271 Posted: Sat Feb 2 20:42:05 1985 Date-Received: Fri, 8-Feb-85 00:24:07 EST References: <423@decwrl.UUCP> Organization: U.Texas Physics Department; Austin, Texas Lines: 34 > Does anyone know the solution y=f(t) for the following differential equation? > > 2 ,, , > t y (t)=K y (0)=K1, y(0)=K2 > ,, , > where K, K1 and K2 are constants, y is the second derivitive, and y is the > first derivitive and t is the independant variable? This is an equation of > the motion of an object under the influence of gravity as a function of time. > > Mike Moroney > ..!decvax!decwrl!rhea!jon!moroney Isn't this an example of a variables-seperable equation? I.e., the solution looks something like this: 2 -2 2 d t*t = K*d y Bit hard to write that on an ascii screen. Oh well, what it says is: "The second differential of t divided by t squared equals K time the second differential of y." Now just integrate twice and plug in the initial conditions. Or have I missed something? -- John L. Templer University of Texas at Austin {allegra,gatech,seismo!ut-sally,vortex}!ut-ngp!lindley "and they called it, yuppy love."