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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