Question: a) If y = $(x + (x)^{\frac{1}{2}}-1)^m , Prove that (x^2-1)y_(n+2)+ (2n+1)xy_(n+1) + (n^2-m^2)y_n= 0$

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$y = (x + √x^2-1)^m$ taking + sign before the radical $∴y_1= m[(x +√x^2-1)^(m-1)].[1+\frac{x}{(√x^2-1)}]$ $= m(x + √x^2-1)^m).\frac{x}{(√x^2-1)} =\frac{my}{(√x^2-1)}$ $√x^2-1 .y_1= my$ Differentiating again w.r.t x, $√x^2-1 .y_2 + x/(√x^2-1) y_1 = my_1$ $(x^2-1)y_2 + xy_1 = m√x^2-1.y_1= m.my = my^2$ $(x^2-1)y_2 + xy_1 - my^2 = 0$ Hence after applying lebnitz’s theorem we get, $(x^2-1)y_(n+2)+ (2n+1)xy_{n+1} + (n^2-m^2)y_n= 0$

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