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General solution of second order differential equaiton.

  1. Feb 19, 2010 #1
    1. The problem statement, all variables and given/known data
    Show that:
    [tex]\theta (\xi) = 1-\frac{1}{6} \xi^2 + \frac{n}{120} \xi^4 +.....[/tex]
    is a general solution to the Lane-Emden equation. (assuming that the above equation converges)


    2. Relevant equations
    Lane-Emden equation:
    [tex]\frac{1}{\xi^2}\frac{d}{d \xi} \left ( \xi^2 \frac{d\theta}{d \xi} \right ) + \theta^n =0 [/tex]


    3. The attempt at a solution
    [tex]\frac{1}{\xi^2}\frac{d}{d \xi} \left ( \xi^2 \frac{d\theta}{d \xi} \right ) + \theta^n = \frac{1}{\xi^2} \left( \xi^2 \frac{d^2 \theta}{d \xi^2} + 2 \xi \frac{d\theta}{d \xi}\right )+ \theta^n =\frac{d^2 \theta}{d \xi^2}+ \frac{2}{\xi} \frac{d\theta}{d \xi}+ \theta^n =0[/tex]
    [tex]\frac{d\theta}{d \xi}=\frac{-1}{3}\xi+\frac{n}{30}\xi^3+...[/tex]
    [tex]\frac{d^2 \theta}{d \xi^2}=\frac{-1}{3}+\frac{n}{10}\xi^2+...[/tex]
    [tex]\frac{2}{\xi}\frac{d\theta}{d \xi}=\frac{-2}{3}+\frac{n}{15}\xi^2+...[/tex]
    [tex]\theta^n=\left( 1-\frac{1}{6} \xi^2 + \frac{n}{120} \xi^4 +.....\right )^n[/tex]
    Plugging them in we get:
    [tex]\frac{1}{\xi^2}\frac{d}{d \xi} \left ( \xi^2 \frac{d\theta}{d \xi} \right ) + \theta^n \left ( \frac{-1}{3}+\frac{n}{10}\xi^2+...\right ) +\left ( \frac{-2}{3}+\frac{n}{15}\xi^2+...\right ) +\left( 1-\frac{1}{6} \xi^2 + \frac{n}{120} \xi^4 +.....\right )^n[/tex]
    Simplifying the 1st 2 terms:
    [tex]=\left ( -1+\frac{n}{6}\xi^2+...\right )+\left( 1-\frac{1}{6} \xi^2 + \frac{n}{120} \xi^4 +.....\right )^n[/tex]
    Well, since the series converges I can just look at the 1st 2 terms in the sums (am I correct?). I need to deal with:
    [tex]\left( 1-\frac{1}{6} \xi^2 + \frac{n}{120} \xi^4 +.....\right )^n[/tex]
    I don't know how to write it out mathematically but I can try to say what I think. After expanding, the first term I'll get is 1, or more precisely, 1^n. The xi^2 term can only come from the product of the first two terms in the series since the product between any other terms will get me a higher power of xi. I tried it out with the first few n's and find that there will be n of the xi^2 term. Therefore:
    [tex]\left( 1-\frac{1}{6} \xi^2 + \frac{n}{120} \xi^4 +.....\right )^n=1^n-\frac{n}{6}\xi^2+...[/tex]
    So theta is indeed the solution (but how is it general? is the question wrong?) Also, how can I write out what I just said more mathematically?
     
  2. jcsd
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