Title: | Mathematics at DEC |
Moderator: | RUSURE::EDP |
Created: | Mon Feb 03 1986 |
Last Modified: | Fri Jun 06 1997 |
Last Successful Update: | Fri Jun 06 1997 |
Number of topics: | 2083 |
Total number of notes: | 14613 |
From: ROLL::USENET "USENET Newsgroup Distributor 27-Feb-1986 2108" 27-FEB-1986 21:19 To: @[.net.math]NEWS.DIS Subj: USENET net.math newsgroup articles Newsgroups: net.math Path: decwrl!dec-rhea!dec-jon!moroney Subject: Re: value of an integral (differential equation) Posted: 27 Feb 86 02:49:55 GMT Organization: Digital Equipment Corporation Can anyone evaluate the following differential equation to the form y=f(x) (i.e. find f(x)) " 2 y y = K y(0)=K y'(0)=K 1 2 K > 0 1 (The " means second derivitive, ' means first derivitive, K, K , K are fixed constants) 1 2 Thanks in advance. -Mike Moroney ..decwrl!dec-rhea!dec-jon!moroney
T.R | Title | User | Personal Name | Date | Lines |
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444.1 | MAPLE can't solve it | METOO::YARBROUGH | Fri Feb 28 1986 09:52 | 15 | |
This is not an easy problem and may not be solvable in elementary functions. MAPLE gives the following 'solution' [int() = integral] : 1/2 y int(---------------------, y) = x + C1, 1/2 1/2 2 (- k + C y) 1/2 y int(- ---------------------, y) = x + C1 1/2 1/2 2 (- k + C y) | |||||
444.2 | If you accept the assumptions... | LATOUR::JMARTIN | Joseph A. Martin | Fri Feb 28 1986 16:01 | 12 |
Y = ((3*SQRT(K1)*K2*X)/2 + SQRT(K1**3)) ** (2/3) assuming that K1 and K2 are the interesting boundary conditions while 2 y"y = K just means that the left hand side is constant but K is not specified. If any of my old calculus students are out there, here is your chance to catch me blowing the chain rule. It's been a while. --Joe |