If \(y=\sin\left(m\sin^{-1}x\right)\), prove that \((1-x^2)y_{n+2}-(2n+1)x\,y_{n+1}+(m^2-n^2)\,y_n=0\).

Answer: First \((1-x^2)y_2-xy_1+m^2y=0\); applying Leibniz's theorem \(n\) times gives the required relation.

Step 1: \(y_1=\cos(m\sin^{-1}x)\cdot\dfrac{m}{\sqrt{1-x^2}}\Rightarrow\sqrt{1-x^2}\,y_1=m\cos(m\sin^{-1}x)\).

Squaring: \((1-x^2)y_1^2=m^2\cos^2(m\sin^{-1}x)=m^2(1-y^2)\).

Step 2: Differentiate w.r.t. \(x\): \((1-x^2)\,2y_1y_2-2x\,y_1^2=-2m^2y\,y_1\). Divide by \(2y_1\):

\[(1-x^2)y_2-xy_1+m^2y=0.\]

Step 3 (Leibniz): \((uv)_n=u_nv+{}^nC_1u_{n-1}v_1+{}^nC_2u_{n-2}v_2+\dots\)

\(\frac{d^n}{dx^n}[(1-x^2)y_2]=(1-x^2)y_{n+2}+n(-2x)y_{n+1}+\frac{n(n-1)}{2}(-2)y_n\)

\(\frac{d^n}{dx^n}[x\,y_1]=x\,y_{n+1}+n\,y_n\), and \(\frac{d^n}{dx^n}[m^2y]=m^2y_n\).

So \((1-x^2)y_{n+2}-2nx\,y_{n+1}-n(n-1)y_n-x\,y_{n+1}-n\,y_n+m^2y_n=0\)

\[\Rightarrow (1-x^2)y_{n+2}-(2n+1)x\,y_{n+1}+(m^2-n^2)y_n=0.\]