$y = \sin [\log (x^2+2x+1)],$ Differentiating w.r.t x, $y_1 = \cos \left [ \log (x^2 +2x+1) \right ] \dfrac {d}{dx} [ \log (x^2 +2x+1)]= \dfrac {\cos [\log (x^2 +2x+1)]}{x^2 +2x+1} \dfrac {d}{dx} (x^2 + 2x +1)$ $y_1 =2 \dfrac {(x+1) \cos [\log (x^2 +2x +1)]} { (x+1)^2}= 2 \dfrac {\cos [\log (x^2 +2x+1)]}{x+1}$ Differentiating w.r.t x, again $y_2=2 \dfrac {(x+1) \left [ -\sin \left [ \log \left ( x^2 +2x+1 \right ) \right ] \frac {d}{dx} \left [ \log \left ( x^2+2x+1 \right ) \right ] - \cos \left [ \log (x^2 +2x+1) \right ] \right ]}{(x+1)^2}$ $y_2 = 2 \dfrac {(x+1 \left [ -y \frac {2}{x+1} \right ])}{(x+1)^2}- \dfrac {1}{x+1} \left [ 2\dfrac{\cos \left [ \log (x^2 +2x+1) \right ]}{x+1} \right ] $ $\ y_2 = - \dfrac {4y}{(x+1)^2}- \dfrac {y_1}{x+1} $ $ (x+1)^2 y_2 + (x+1)y_1 + 4y =0 $ Differentiating this n times using Lebnitz Theorem, $(x+1)^2 y_{n+2}+ 2(x+1)n y_{n+1}+2 \dfrac {n (n-1)}{1\times 2} y_n + (x+1)y_{n+1}+ (1\times n) y_n + 4y_n =0 $ $(x+1)^2 y_{n+2}+ [2n (x+1)+x+1] y_{n+1}+ [n^2 -n+n+4]y_n=0 $ $(x+1)^2 y_{n+2}+ (2n+1)(x+1)y_{n+1} + (n^2 +4) y_n =0$

Hence Proved.