Differential equations

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.

(edited 12 years ago)

Just skimming before I go to bed, and what you have so far seems good to me. :tongue:

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). :biggrin:

Spoiler

(edited 12 years ago)

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

Thanks.

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