$y = \sin [\log (x^2+ 2x+1)] = \sin [\log (x+1)^2 ] $

$\\ y = \sin [2\log(x+1) ] ....(A)$

$ \\ \text{Differentiating implicitly w.r.t .x, using chain rule}$

$y_1 = \cos [2\log(x+1)] [2 \log (x+1)]' = 2 \dfrac{\cos [2\log(x+1)]}{x+1}(x+1)' $

$ y_1 = 2 \dfrac{\cos [2\log(x+1)]}{x+1}$

$(x+1)y _1 = 2 \cos [2\log(x+1)]$

$ \text {Differentiating implicitly again w.r.t x, using product and chain rule, }$

$ \\ (x+1)y_2 + y_1 = 2 [-\sin[2\log(x+1)]] \times [2\log(x+1)]'$

$(x+1)y_2 + y_1 = -4 \dfrac{\sin [2\log(x+1)] }{x+1} (x+1)'$

$(x+1)^2 y_2 + (x+1)y_1 = - 4 y ....[from (A)] $

$\\ (x+1)^2 y_2 + (x+1)y_1 + 4y = 0 $

$\text{Differentiating n times using Lebnitz Theorem,} $

$\\ (x+1)^2 y_{n+2} + 2(x+1)\times (n) y_{n+1} + 2 \times [\dfrac{n(n-1)}{2}] y_n + (x+1)y_{n+1} + ny_n +4y_n = 0$

$\text{Combining} \quad y_{n+1} \quad and \quad y_n \quad terms,$

$ \\ (x+1)^2 y_{n+2} + [2n (x+1) + (x+1)] y_{n+1} + [n(n-1) + n+4]y_n = 0 \\ $

$Hence, \\ \bbox[10pt,yellow]{( x+1)^2 y_{n+2} + (2n+1)(x+1)y_{n+1} + (n^2 +4) y_n = 0 }$