Question

$y=e^{tan^{-1}x}\\ P.T (1+x^2)y_{n+2}+[2(n+1)x-1]y_{n+1}$

Solution

$y=e^{tan^{-1}x}\\ y_1 =\dfrac{dy}{dx}= e^{tan^{-1}x}.\dfrac{1}{1+x^2} \\ =\dfrac{y}{1+x^2} \\ \therefore y_1= \dfrac{y}{1+x^2} \\ y_1(1+x^2)=y$

Differentiating once again

$y_2 (1+x^2) +y_1.2x=y_1\\ y_2 (1+x^2) +y(2x-1)=0$

Applying Lebnitz theorem to above equation

$(1+x^2) y_{n+2}+y_{n+1}.2x.n+y_n.\dfrac{n(n-1)}{2}.2+y_{n+1}(2x-1)+y_n(2).n =0\\ (1+x^2) y_{n+2} +y_{n+1}(2x.n+2n-1)+y_n(n^2-n+2n)=0\\ (1+x^2) y_{n+2} +[2(n+1)x-1]y_{n+1}+n(n+1)y_n=0 \ is\ proved$