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If $ sin[log(x^2+2x+1)]$ ,prove that $(x+1)^2 y_{n+2}+(2n+1)(x+1) y_{n+1}+(n^2+4) y_n=0 $
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y$ =sin⁡[log⁡(x^2+2x+1) ] \;=\; sin⁡[log⁡(x+1)^2 ] $

Differentiating y w.r.t.x,we get-

$y_1=cos⁡[log(x+1)^2 ] \dfrac{1}{(x+1)^2} 2(x+1) \;=\; \dfrac{2 cos⁡[log(x+1)^2 ]}{x+1} \; \; \; \ldots (i) $

Again differentiating y w.r.t.x,

$ y_2= \dfrac{(x+1){-2 sin⁡[log⁡(x+1)^2 ] 1/(x+1)^2 2(x+1)}-2 cos⁡[log⁡(x+1)^2 ] (1)}{(x+1)^2} \\ \; \\ \; \\ \therefore y_2= \dfrac{-4 sin⁡[log(x+1)^2 ]-2 cos⁡[log⁡(x+1)^2 ]}{(x+1)^2 } \\ \; \\ \; \\ y_2= \dfrac{-4y-(x+1)y_1 }{(x+1)^2} \; \; \; \; \ldots From \; (i) \\ \; \\ \; \\ \therefore (x+1)^2 y_2+(x+1) y_1+4y=0 $

Differentiating above equation w.r.t.x n times,

Using Leibnitz' s Theorem,

$ (UV)_n= n_{C_0} U_n V + n_{C_1} U_{n-1} V_{1} + n_{C_2} U_{n-2} V_{2} + \ldots + n_{C_n} U V_{n} + \\ \; \\ \; \\ \therefore (x+1)^2 y_{n+2}+2n(x+1) y_{n+1}+2(1) \dfrac{n(n-1)}{2} y_n+(x+1) y_{n+1} +n(1) y_n+4y_n=0 \\ \; \\ \; \\ \therefore (x+1)^2 y_{n+2}+(2n+1)(x+1) y_{n+1}+(n^2-n+n+4) y_n=0 \\ \; \\ \; \\ i.e. \; \; (x+1)^2 y_{n+2}+(2n+1)(x+1) y_{n+1}+(n^2+4) y_n=0 $

Hence Proved.

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