Question

$x=\sin ty=\cos mt$
Prove $\left(1-x^2\right)y_{n+2}-\left(2n-1\right)y_{n+1}-\left(n^2-m^2\right)y_n=0$

Answer 1

$x=\sin ty=\cos mt$

$t={\sin }^{-1}\left(x\right)$ $y=\cos \left(m{\sin }^{-1}\right(x\left)\right)$

$(1-x^2)y_{n+2}-(2n+1)x{y}_{n+1}+(m^2-n^2)y_n=0$

$\frac{dy}{dx}=-\sin \left(m{\sin }^{-1}\right(x\left)\right)=\frac{m}{\sqrt{1-x^2}}$

$\sqrt{1-x^2}\frac{dy}{dx}=-m\sin \left(m{\sin }^{-1}x\right)$

$(1-x^2){\left(\frac{dy}{dx}\right)}^2=-m^2{\left[\sin \right(m{\sin }^{-1}x\left)\right)]}^2$

$=m^2[1-{\cos }^2(m{\sin }^{-1}x\left)\right]$

$(1-x^2){\left(\frac{dy}{dx}\right)}^2=m^2[1-y^2]$ $|\begin{array}{c}\frac{dy}{dx}=y_1\\ \frac{d^{2}y}{{dx}^2}=y_2\end{array}$

$(1-x^2)2y_1{y}_2+y_1^2(-2x)=m^2(-2y)y_1$

$(1-x^2)y_2-x{y}_1=-m^{2}y$

$\therefore (1-x^2)y_2-x{y}_1+m^{2}y=0$

$(1-x^2)y_{m+2}+m(-2x{)}_{ym+1}\ne \frac{m(n-1)}{2}.(-2)y_m$

$x{y}_{x+1}-x{y}_m-m^2{y}_n=0$

$(1-x^2)y_{n+2}-x(2x++1)Y_{n+1}+(n^2+n-n+m^2)yx=0$

$(1-x^2)y_{m+2}-(2n+1)x{y}_{x+1}+(m^2-n^2)yx=0$