| 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 |
|---|---|---|---|---|---|
| 444.1 | MAPLE can't solve it | METOO::YARBROUGH | Fri Feb 28 1986 12:52 | 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)
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| 444.2 | If you accept the assumptions... | LATOUR::JMARTIN | Joseph A. Martin | Fri Feb 28 1986 19:01 | 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
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