$\text {if } x =\cos \Big [ \log \big (y^{1/m} \big) \Big ] \ \arccos x = \dfrac{1}{m}\log y \\log y = m \arccos x $ $\ \text{Differentiating this Implicitly,} \ \dfrac{y_1}{y} = \dfrac{-m}{\sqrt{1-x^2}} \ y_1 \sqrt{1-x^2} = -my ...(A)$ $\text{Differenting this implicitly and using product and chain rule,} \ y_2\sqrt{1-x^2} + \dfrac{y_1}{2\sqrt{1-x^2}} (\sqrt{1-x^2})'= -my_1 \ y_2\sqrt{1-x^2} - \dfrac{xy_1}{\sqrt{1-x^2}} = -my_1 $ $\ \text{Multiplying } \sqrt{1-x^2} \quad throughout, \ y_2(1-x^2) -xy_1 = -my_1\sqrt{1-x^2}$ $\ y_2 (1-x^2) -xy_1 = -m(-my) ...(From (A)) \ y_2 (1-x^2) -xy_1 = m^2 y $ $\ y_2 (1-x^2) -xy_1-m^2 y = 0 \ \text{Differentiating n times using Lebnitz Theorem,} $ $(1-x^2)y_{n+2} + (-2x)\times (n)y_{n+1} + (-2)[\dfrac{n(n-1)}{2}]y_n -xy_{n+1}+(-1)\times ny_{n} -m^2 y_n = 0\ \text{Combining } y_{n+1} \: and \: y_n \: terms,$ $(1-x^2)y_{n+2} + (-2nx-x)y_{n+1} + [-n^2+n-n-m^2]y_n = 0 $ $\bbox[3pt, yellow] {\therefore, (1-x^2)y_{n+1}-(2n+1)xy_{n+1}-(m^2+n^2)y_n = 0}$