Rayleigh's differential equation

atrus_ovis

In Rayleigh's DE :
http://www.wolframalpha.com/input/?i...ntial+equation

What does mu stand for?

dextercioby

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

atrus_ovis

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

dextercioby

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

atrus_ovis

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?