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

Consider the given equation

$x=\sin t$ …….. (1) $y=\cos mt$ ……… (2)

By equation (1),

Given, $x=\sin t$ then $t={\sin }^{-1}x$ put in equation (2)

We get, $y=\cos \left(m{\sin }^{-1}x\right)$ …….. (3)

Differentiating equation (3) with respect to x and we get,

$y_1=\frac{dy}{dx}=-\sin \left(m{\sin }^{-1}x\right).\frac{m}{\sqrt{1-x^2}}$

$y_1\sqrt{1-x^2}=-m\sin \left(m{\sin }^{-1}x\right)$

Squaring both side,

$y_1\sqrt{1-x^2}=-m\sin \left(m{\sin }^{-1}x\right)$

$\Rightarrow (1-x^2){y_1}^2=m^2{\left[\sin \left(m{\sin }^{-1}x\right)\right]}^2$

$\Rightarrow (1-x^2){y_1}^2=m^2[1-{\cos }^2\left(m{\sin }^{-1}x\right)]$

$\Rightarrow (1-x^2){y_1}^2=m^2(1-y^2)..........by\left(3\right)$

Again, Differentiating with respect to x and we get,

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

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

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

Differentiating this equation $n$ times by Leibnitz rule,

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

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

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

Hence proved.