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Topic: solve 5 (Read 704 times) |
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Grimbal
wu::riddles Moderator
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Re: solve 5
« Reply #1 on: Jul 14th, 2007, 1:41am » |
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i.e.
Solve the differential equation
(tan x + m sin y) dy = (sin y - m tan x) cos y dx
where m is a constant.
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Aryabhatta
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Re: solve 5
« Reply #2 on: Jul 14th, 2007, 11:26am » |
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tony123, a request.
Can we please have more descriptive titles?
Also, why don't you attach the image to the post? I am not sure if the link you gave will be even valid a year from now.
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Sameer
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Pie = pi * e
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Re: solve 5
« Reply #3 on: Jul 14th, 2007, 12:00pm » |
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I gave this request earlier with no effect. Maybe we need some moderator intervention!!!
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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tony123
Junior Member
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Re: solve 5
« Reply #4 on: Jul 15th, 2007, 12:49am » |
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I am sorry and I apologize
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Michael Dagg
Senior Riddler
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Re: solve 5
« Reply #5 on: Aug 1st, 2007, 4:15pm » |
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If m=0 you can separate the variables and completely solve
the problem (which I shall leave up to you).
Otherwise, the situation is not that simple. If you view the
problem as a phase plane problem, then something interesting
happens. The differential equation then becomes a 2-dimensional
system
x'= tan(x) + m sin(y)
y'= (sin(y) - m tan(x)) cos(y) .
Solving the original problem amounts to finding a function
F(x,y) so that F_x(x,y) x' + F_y(x,y) y' = 0, or, in other words,
F is constant along solutions of the system given above.
There is an obvious solution of this system (x(t),y(t))=(0,0),
which is a rest point at the origin. All solutions starting nearby
the origin will approach the origin as t goes to negative infinity.
You can easily check this with a computer program.
For the case m=0 their approach will have a definite tangent
in the limit; but, for the case m not zero the solutions will
spiral around the origin and not have a definite tangent in
the limit. This is pretty good evidence that leads me to
believe that there is unlikely a closed form formula for a
function F that is constant on each such spiral orbit -- but it
is not impossible!!!
One thing is for sure is that the function F cannot be continuously
defined at the origin. This is true for all cases of m, even
m=0 . If F could be continuously defined then it would have
to be constant in some neighborhood of the origin exactly
because F is constant on solutions, they fill up the
neighborhood, and they all limit at the same point. So, at
best, the "solution" we might find for the original problem
will be singular at the origin.
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| « Last Edit: Aug 1st, 2007, 4:18pm by Michael Dagg » |
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Regards,
Michael Dagg
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Michael Dagg
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Re: solve 5
« Reply #6 on: Aug 4th, 2007, 7:25pm » |
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As I said, it is not impossible! I have found an integrating
factor for this ODE. If you multiply the ODE by cos(x) to
lose the tangent function therein, and then use the
integrating factor
p = 1/(1 - (cos(x)cos(y))^2)
the ODE then becomes exact!
The integrals are a mess but it is now possible to find a
closed form formula in terms of elementary functions!
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| « Last Edit: Aug 4th, 2007, 10:45pm by Michael Dagg » |
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Regards,
Michael Dagg
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