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From: leimkuhl@uiucdcsp.UUCP
Newsgroups: net.math
Subject: Re: help solve this?
Message-ID: <9600002@uiucdcsp.UUCP>
Date: Sun, 10-Feb-85 15:31:00 EST
Article-I.D.: uiucdcsp.9600002
Posted: Sun Feb 10 15:31:00 1985
Date-Received: Tue, 12-Feb-85 06:13:42 EST
References: <1351@hao.UUCP>
Lines: 52
Nf-ID: #R:hao:-135100:uiucdcsp:9600002:000:1187
Nf-From: uiucdcsp!leimkuhl    Feb 10 14:31:00 1985




   This is actually a rather interesting problem.


   If we try the approach suggested above, we get the rather
nasty recurrence relation:

	a1*a0**2 - A*a0 + B = 0
	a2*a0**2 + 2*a0*a1**2 - A*a1 - B = 0

        Sum(i=0,n; a(n-i+1)*Sum(j=0,i; aj*a(i-j)))  -  A*an = 0
                for n=2,3,..

   This is too hard to solve I think.


   

   Let L(y) = y'*y**2 - Ay

   The homogeneous equation L(y)=0 has the relatively simple solution

           y**2 = Ax + K 

   --a family of parabolas.

   When we progress to the seemingly still simple problem L(y)=B, we
   get the solution

           (u**2)/2 - 2*B*u + (B*B)*log u = k + x*A**3,

   where u = Ay + B


   This is already enormously complicated and impossible to solve 
   explicitly for u.


   Now the given problem is to solve L(y) = B - Bx  which obviously
   is not going to have a simple solution (since L(y)=B is just a
   special case).

   In fact, we can rule out simple forms of a solution involving 
   complex exponentials and (what is the same thing, trigonemetric
   functions).  The method of judicious guessing is not likely to
   yield anything of value.

   Some other suggestions?  

-Ben Leimkuhler