Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site ho95b.UUCP Path: utzoo!watmath!clyde!cbosgd!cbdkc1!desoto!packard!hoxna!houxm!ho95b!ran From: ran@ho95b.UUCP (RANeinast) Newsgroups: net.math Subject: Re: Solution to the NEW weighty problem. Message-ID: <307@ho95b.UUCP> Date: Thu, 7-Feb-85 15:38:25 EST Article-I.D.: ho95b.307 Posted: Thu Feb 7 15:38:25 1985 Date-Received: Sat, 9-Feb-85 07:36:13 EST Organization: AT&T-Bell Labs, Holmdel, NJ Lines: 26 # >Sorry, folks, but I made a mistake. The equation I meant is as follows: > > ,, , > y (t) = K/(y^2) y (0)=K1, y(0)=K2. > >Mike Moroney This one's almost as easy as the mistake. Multiply both sides by y' y'*y''=K*(y^(-2))*y' Both sides are now perfect differentials. Integrate once and get y' in terms of y. Now look it up in an integral table. -- ". . . and shun the frumious Bandersnatch." Robert Neinast (ihnp4!ho95c!ran) AT&T-Bell Labs