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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