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Title: solve 5 Post by tony123 on Jul 12th, 2007, 4:14pm http://endeavor.macusa.net/mathpropress/gif/00/5/00591.gif |
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Title: Re: solve 5 Post by Grimbal on Jul 14th, 2007, 1:41am 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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Title: Re: solve 5 Post by Aryabhatta on Jul 14th, 2007, 11:26am 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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Title: Re: solve 5 Post by Sameer on Jul 14th, 2007, 12:00pm I gave this request earlier with no effect. Maybe we need some moderator intervention!!! |
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Title: Re: solve 5 Post by tony123 on Jul 15th, 2007, 12:49am I am sorry and I apologize |
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Title: Re: solve 5 Post by Michael_Dagg on Aug 1st, 2007, 4:15pm 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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Title: Re: solve 5 Post by Michael_Dagg on Aug 4th, 2007, 7:25pm 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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