Let y=sin(n*sin^-1x), n>0.Show that: (1 - x^2)*y" - xy' + n^2y=0

Is this Chebyshev's equation: (1 - x^2)*y" - xy' + n^2y=0   ???????

y = sin (n * arcsin x)   n > 0

y '  =  cos (n*arcsin x) * [ n  (1 - x^2)^(-1/2)]

y ''  = -sin(n*arcsin x) * [ n  (1 - x^2)^(-1/2)]* [ n  (1 - x^2)^(-1/2)]  + cos(n * arcsinx)* [n * x * (1 - x^2)^(-3/2) ] =

-sin(n*arcsin x)* [n^2 (1- x^2)^-1] + cos (n*arcsinx) * [ nx * (1- x^2)^(-3/2) ]  =

[n (1-x)^(-3/2)] * [ xcos(n*arcsinx  - nsin(n*arcsinx)(1-x^2)^(1/2) ]

So....

(1 - x^2) y ''  - xy '  + n^2y    =

(1-x^2) [n (1-x)^(-3/2)] [ xcos(n*arcsinx)- nsin(n*arcsinx)(1-x^2)^(1/2)] -

x [cos (n*arcsin x)] * [ n  (1 - x^2)^(-1/2)]  + n^2 [sin (n * arcsin x)]  =

[ n (1 -x)^(-1/2) ] [ xcos(n*arcsinx)  - nsin(n*arcsinx)(1-x^2)^(1/2) ] 

- x [cos (n*arcsin x)] * [ n  (1 - x^2)^(-1/2)]  + n^2 [sin (n * arcsin x)]   =

[ xcos(n*arcsinx)] *  [ n (1 -x)^(-1/2) ] -

[ n (1 -x)^(-1/2) ] * [ nsin(n*arcsinx)(1-x^2)^(1/2)] -

[x cos (n*arcsin x)] * [ n  (1 - x^2)^(-1/2)]  + n^2 [sin (n * arcsin x)]   =

n^2 [sin (n * arcsin x)]  -

 n * (1 -x)^(-1/2)  * [ nsin(n*arcsinx)] * (1-x^2)^(1/2)  =

n^2 [sin (n * arcsin x)]   -  [ n* n * (1 -x)^(-1/2) * (1 -x)^(1/2) * sin(n*arcsinx) ] =

n^2 [sin (n * arcsin x)]  - n^2 [ sin(n * arcsinx)]  =

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