1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Trigonometric form of Legendre equation
  1. Legendre Equation (Replies: 2)

  2. Legendre's equation (Replies: 2)

  3. Trigonometric equation (Replies: 23)

  4. Trigonometric equation (Replies: 6)

Loading...