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[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

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# Homework Help: Trigonometric form of Legendre equation

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