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- Newsgroups: sci.math
- Subject: Re: ODE problem...
- Message-ID: <a_rubin.722192494@dn66>
- From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
- Date: 19 Nov 92 17:01:34 GMT
- References: <1992Nov17.205237.21447@athena.mit.edu>
- Organization: Beckman Instruments, Inc.
- Nntp-Posting-Host: dn66.dse.beckman.com
- Lines: 53
-
- In <1992Nov17.205237.21447@athena.mit.edu> frisch1@athena.mit.edu (Jonathan Katz) writes:
-
- >The following ODE problem came up recently.
- >I know how to solve it by the power series method, but was wondering
- >if anyone could figure out an easier way of solving it (maybe a nice
- >substitution?).
- >(x and y are functions of t, a is a constant)
- >x'=(a)(x)cost+(a)(y)sint
- >y'=(a)(x)sint-(a)(y)cost.
- >It may be worth noting that, viewing this as a matrix problem:
- >x'=Ax,
- >A^2=I.
-
- >I don't know if that helps.
-
- I used a completely different approach than Robert. If you write:
-
- z = x + i y, then the equation becomes:
-
- i t _
- z' = E z
-
- Playing with this for a while, and guessing that solutions z exist which
- are a sum of only 2 expontials in t, we can derive the general solution:
-
- Let c be one of the square roots of (a^2-1/4).
-
- C(t) = / cosh(|c| t) , c real != 0
- | 1 , c = 0
- \ cos(|c| t) , c pure imaginary
-
- S(t) = / sinh(|c| t)/|c| , c real != 0
- | t , c = 1
- \ sin(|c| t)/|c| , c pure imaginary
-
-
- x0(t) = C(t) cos(t/2) + S(t) (a cos(t/2) + sin(t/2)/2)
- x1(t) = -C(t) sin(t/2) + S(t) (a sin(t/2) + cos(t/2)/2)
-
- y0(t) = C(t) sin(t/2) + S(t) (a sin(t/2) - cos(t/2)/2)
- y1(t) = C(t) cos(t/2) + S(t) (-a cos(t/2) + sin(t/2)/2)
-
- Then x=C0 x0 + C1 x1 and y=C0 y0 + C1 y1 form the general solution.
-
-
-
-
-
- --
- Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
- 216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
- My opinions are my own, and do not represent those of my employer.
- My interaction with our news system is unstable; please mail anything important.
-