Consider the Chebyshev polynomial of the first kind $$ (1-x^2)y'' - xy' + n^2y = 0 , n \in \mathbb{N}. $$ Use the substitution $ x=\cos\theta $ and show that the transformed ODE has solutions $y_1 = \cos (n\theta)$ and $y_2 = \sin (n\theta)$.
I have tried substituting in $\cos \theta $ for $x$ and then substituting in the given solutions for $y$, but I have been unable to show it is equal to zero.
The substitution yields $$ (1-\cos^2 \theta )y'' - \cos (\theta)y' + n^2y$$ and upon substituting in the respective derivatives of $y_1$, I get the left had side to be $$ (1- \cos^2 \theta )(-n^2 \cos (n \theta )) - \cos (\theta )(-n \sin (n\theta )) + n^2\cos (n\theta)$$
which I am struggling to show equals zero.