y$^{\dfrac{1}{m}} + y^{\dfrac{-1}{m}} \;=\; 2x \\ \; \\ \; \\ \therefore y^{\dfrac{1}{m}} + \dfrac{1}{y^{\dfrac{1}{m}}} \;=\; 2x \\ \; \\ \; \\ \therefore y^{\dfrac{2}{m}} + 1 = 2xy^{\dfrac{1}{m}} \\ \; \\ \; \\ \therefore y^{\dfrac{2}{m}} - 2xy^{\dfrac{1}{m}} + 1 =0 \\ \; \\ \; \\ \therefore y^{\dfrac{1}{m}} \;=\; \dfrac{ -(-2x) \pm \sqrt{4x^2-4(1)(1)} }{2(1)} \;=\; \dfrac{ 2x \pm \sqrt{4x^2-4} }{2} \\ \; \\ \; \\ y^{\dfrac{1}{m}} \;=\; \dfrac{ 2\Bigg[ x \pm \sqrt{x^2-1} \Bigg] }{2} \;=\; x \pm \sqrt{x^2-1} \\ \; \\ \; \\ \therefore y \;=\; \bigg( x \pm \sqrt{x^2-1} \bigg)^m \; \; \; \; \; \; \ldots (i) $

Diff. w.r.t.x the above equation (i),

$ \therefore y_1 \;=\; m \bigg( x \pm \sqrt{x^2-1} \bigg)^{m-1} \Bigg( 1 \pm \dfrac{1}{2\sqrt{x^2-1}} (2x) \Bigg) \\ \; \\ \; \\ \therefore y_1 \;=\; m \bigg( x \pm \sqrt{x^2-1} \bigg)^{m-1} \Bigg( 1 \pm \dfrac{x}{\sqrt{x^2-1}} \Bigg) \\ \; \\ \; \\ \therefore y_1 \;=\; m \bigg( x \pm \sqrt{x^2-1} \bigg)^{m-1} \Bigg( \dfrac{ \sqrt{x^2-1} \pm x }{\sqrt{x^2-1}} \Bigg) \\ \; \\ \; \\ \therefore y_1 \;=\; \dfrac{ m \bigg( x \pm \sqrt{x^2-1} \bigg)^{m-1} }{\sqrt{x^2-1}} \\ \; \\ \; \\ \therefore y_1 \sqrt{x^2-1} \;=\; my \; \; \; \ldots (i) $

Squaring on both sides,

$ y_1^2 (x^2-1)^2 \;=\; m^2y^2 $

Differentiating w.r.t.x,

$ \therefore (x^2-1)2y_1y_2 + y_1^2(2x) = m^2 2yy_1 \\ \; \\ \; \\ \therefore 2(x^2-1)y_2 + 2xy_1 - 2m^2y =0 \\ \; \\ \; \\ (x^2-1)y_2 + xy_1 - m^2y =0 \; \; \; \ldots (ii) $

Diff. equation (ii) n times by using Leibnitz’s theorem,

$ \therefore (x^2-1)y_{n+2} + n(2x)y_{n+1} + \dfrac{n(n-1)}{2} .(2) y_n + xy_{n+1} + ny_n - m^2y_n =0 \\ \; \\ \; \\ \therefore (x^2-1)y_{n+2} + x(2n+1)y_{n+1} + (n^2-n+n-m^2) y_n =0 \\ \; \\ \; \\ \therefore (x^2-1)y_{n+2} + (2n+1)xy_{n+1} + (n^2-m^2)y_n =0 \; \; \; \; \; \ldots Hence \; Proved. $