Trigonometric form of Legendre equation

[Jesssa]

hey,

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

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

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

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

im just a little paranoid this seems a bit straightforward for a past test question

 

[LCKurtz]

hey,

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

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

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

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

im just a little paranoid this seems a bit straightforward for a past test question

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