We have

$y^{1/m} + y^{-1/m} = 2x$

$\therefore y^{2/m} - 2xy^{1/m} + 1 = 0$

This is quadratic equation in y1/m

$y^{1/m} = \dfrac {2x \pm \sqrt{4x^2 - 4}}{2} = x\pm \sqrt{x^2 - 1}$

$\therefore y = \left ( x \pm \sqrt{x^2 -1} \right ) ^m$

Taking the + sing before the radical

$\begin {align*} \therefore y_1 &= m \left ( x + \sqrt{x^2 - 1} \right )^{m-1} \left ( 1+ \dfrac {x}{\sqrt{x^2 - 1}} \right ) = m \left ( x + \sqrt{x^2 - 1} \right )^m \dfrac {1}{\sqrt{x^2 -1}} \\ &= \dfrac {my}{\sqrt{x^2-1}} \end{align*}$

$\therefore \sqrt{x^2 - 1} \ y_1 = my$

Differentiating w.r.t. to x

$\sqrt{x^2 - 1} y_2 + \dfrac {x}{\sqrt{x^2 - 1}} y_1 = my_1$

$(x^2 - 1) y_2 + xy_1 = m^2 y$

$(x^2 - 1) y_2 + xy_1 - m^2 y=0$

Applying Leibnitz Theorem

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

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