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Homework Help: Trigonometric form of Legendre equation

  1. Apr 15, 2012 #1
    hey,

    [itex](1-{{x}^{2}}){{y}^{''}}-2x{{y}^{'}}+n(n+1)y=0,\,\,\,\,\,-1\le x\le 1[/itex]

    to convert the legendre equation y(x) into trig form y(cos[itex]\theta[/itex]) is it simply, set x=cos[itex]\theta[/itex] then

    [itex](1-{{\cos }^{2}}\theta ){{y}^{''}}-2{{y}^{'}}\cos \theta +n(n+1)y=0[/itex] for [itex]-\pi \le x\le \pi[/itex]

    [itex]{{\sin }^{2}}\theta {{y}^{''}}-2{{y}^{'}}\cos \theta +n(n+1)y=0[/itex] ?

    im just a little paranoid this seems a bit straightforward for a past test question
     
  2. jcsd
  3. Apr 15, 2012 #2

    LCKurtz

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    The problem is you are glossing over what ##y'## means. In the first equation it is ##\frac{dy}{dx}## and in the second it is ##\frac{dy}{d\theta}##. They aren't the same thing. You need the chain rule. You can see the correct formulas here:
    http://mathworld.wolfram.com/LegendreDifferentialEquation.html
     
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