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Rayleigh's differential equation 
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#1
Dec2912, 08:28 AM

P: 101




#2
Dec2912, 09:13 AM

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P: 11,896

It's a mere real positive parameter, just like in other famous ODE's, for example the Bessel differential equation and <n>.
http://www.wolframalpha.com/input/?i...ntial+equation 


#3
Dec2912, 09:44 AM

P: 101

Well, i am asked to numerically solve it and produce a phase diagram.
Should its value be given to me? 


#4
Dec2912, 10:00 AM

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HW Helper
P: 11,896

Rayleigh's differential equation
I guess it should, so you're free to choose any value you want: Take [itex] \mu =1 [/itex] and solve it numerically.



#5
Dec3012, 08:28 AM

P: 101

You're right , it was supposed to be given.
Rayleigh's DE is [itex]y''\mu y' + \frac{\mu (y')^3}{3} + y = 0[/itex] By rearranging it to a system of DEs, you get [tex] y_1 = y , y_1' = y_2 \\ y_2' = \mu y_2  \frac{\mu (y_2)^3}{3}  y_1 [/tex] So i have only the derivative of y2 , i.e. the 2nd derivative of y1. Since i don't have an analytical description of y2 , how do i compute it with specific parameters, according to the numerical method. For example, for the classic Runge Kutta method,where f = y' [tex] k_1 = hf(x_n,y_n) = hy_2(n)\\ k_2 = hf(x_n + 0.5h,y_n + 0.5k_1) = ? [/tex] I should numerically approximate the intermmediate values as well? 


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