Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: Notesfiles $Revision: 1.6.2.17 $; site uiucdcsp.UUCP Path: utzoo!watmath!clyde!burl!ulysses!mhuxr!ihnp4!inuxc!pur-ee!uiucdcsp!leimkuhl 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