[net.math] Solution to the NEW weighty problem.

ran@ho95b.UUCP (RANeinast) (02/07/85)

#

>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