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    I have the ODE  \dfrac{1}{\sin\theta} \dfrac{d}{d \theta} \left(\sin\theta \dfrac{dg}{d \theta} \right) + \Lambda g(\theta) = 0 , and using the substitution  x=\cos \theta , I have to show that it equals the Legendre's equation  (1-x^2) \dfrac{d^2g}{dx^2}-2x\dfrac{dg}{dx} + \Lambda g = 0 .
    What I did was this:
    Rewrote the ODE as  \dfrac{1}{\sin \theta} \cos \theta \dfrac{dg}{d\theta} + \dfrac{d^2g}{d \theta^2} + \Lambda g = 0 ,  \dfrac{dx}{d\theta} = -\sin\theta \Rightarrow -\dfrac{d\theta}{dx} = \dfrac{1}{\sin\theta} then subbed this into the ODE to get  -x\dfrac{d\theta}{dx} \dfrac{dg}{d\theta} + \dfrac{ d^2g}{d\theta^2} +\Lambda g=0. Using the chain rule \dfrac{d^2g}{d\theta^2} = \dfrac{d^2g}{dx^2} \left(\dfrac{dx}{d\theta}\right)  ^2 = (1-x^2)\dfrac{d^2g}{dx^2} and ended up with  (1-x^2) \dfrac{d^2g}{dx^2}-x\dfrac{dg}{dx}+\Lambda g = 0 . Just want to know where did I go wrong on this? Thanks.
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    Just skimming before I go to bed, and what you have so far seems good to me.

    However, in the first equation you have g(\theta) and the second just g?
    Is it just me, or something is missing there (in your chain rule maybe).

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    \displaystyle \frac{d^2 g(\arccos(x))}{dx^2} = \frac{d^2g}{d\theta^2} \left( \frac{-1}{\sqrt{1 - x^2}} \right)^2 + \frac{dg}{d\theta} \left(\frac{-x}{(1 - x^2)^{3/2}} \right)

    which of course has the more useful form of

    \displaystyle \frac{d^2g}{dx^2} = \frac{d^2g}{d\theta^2}\frac{1}{ \sin^2(\theta)} - \frac{dg}{dx} \frac{dx}{d\theta} \frac{ \cos(\theta) }{\sin^3( \theta)}

    I leave it to you to do it the reverse way in which you started.
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    Your mistake is in the chain rule.

    \frac{d^2g}{d\theta^2}=\frac{d}{  d\theta}\left(\frac{dg}{d\theta}  \right)=\frac{d}{dx}\left(\frac{  dg}{dx}\cdot\frac{dx}{d\theta} \right)\frac{dx}{d\theta}\not =\frac{d^2g}{dx^2}\left(\frac{dx  }{d\theta}\right)^2
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    Thanks.
 
 
 
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