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