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- From: israel@unixg.ubc.ca (Robert B. Israel)
- Newsgroups: sci.math
- Subject: Re: ODE problem...
- Date: 23 Nov 92 22:51:38 GMT
- Organization: The University of British Columbia
- Lines: 41
- Message-ID: <israel.722559098@unixg.ubc.ca>
- References: <1992Nov20.184102.14068@athena.mit.edu>
- NNTP-Posting-Host: unixg.ubc.ca
-
- In <1992Nov20.184102.14068@athena.mit.edu> frisch1@athena.mit.edu (Jonathan Katz) writes:
-
- >T
- >Lines: 30
-
- >>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.
-
- >Hi, I was the one to originally pose this problem, and I'm glad to see the
- >interest it has sparked.
- >Thanks to all those who sent email.
-
- >However, while playing with this system, I came upon a neat 'trick' which no one
- >has yet pointed out.
-
- >multiply the first equation by y and the second by x. Then add the two equations
- >together, getting:
- >yx'+xy'= xy(acos(t)+asin(t))
- >letting f=xy, note that this simplifies to:
- >f'=f(acos(t)+asin(t)) which is easily solvable.
-
- >Now subtract the two equations, and divide the resulting equation by y^2. Then
- >let g=x/y, and the equation becomes:
- >g'=some function of g (sorry I don't have my work in front of me, but it's
- >simple enough to carry out)
-
- Huh? y x' + x y' = a (x^2 + y^2) sin(t)
- On the other hand, y x' - x y' = a (2 x y cos(t) + (y^2 - x^2) sin(t))
- so g' = a(2 g cos(t) + (1 - g^2) sin(t)) (which isn't particularly pleasant
- to solve).
-
- --
- Robert Israel israel@math.ubc.ca
- Department of Mathematics or israel@unixg.ubc.ca
- University of British Columbia
- Vancouver, BC, Canada V6T 1Y4
-