$\mathrm{If} \mathrm{y}=\sin \left[\log \left({\mathrm{x}}^2+2\mathrm{x}+1\right)\right] \mathrm{then} {\left(\mathrm{x}+1\right)}^2{\mathrm{y}}_2+\left(\mathrm{x}+1\right){\mathrm{y}}_1=$

$$\begin{gathered} \mathrm{y}=\sin \left[\log ({\mathrm{x}}^2+2\mathrm{x}+1)\right] \\ =\sin \left[\log {(\mathrm{x}+1)}^2\right] \\ =\sin \left[2 \log (\mathrm{x}+1)\right] \end{gathered}$$

$$\begin{gathered} {\mathrm{y}}^1=\cos \left(2\log \left(\mathrm{x}+1\right)\right)\frac{2}{\mathrm{x}+1} \\ (\mathrm{x}+1){\mathrm{y}}^1=2\cos \left(2\log (\mathrm{x}+1)\right) \\ (\mathrm{x}+1){\mathrm{y}}^{11}+{\mathrm{y}}^1=\frac{-4\sin \left(2\log \right(\mathrm{x}+1\left)\right)}{(\mathrm{x}+1)} \\ {(\mathrm{x}+1)}^2 {\mathrm{y}}^{11}+(\mathrm{x}+1) {\mathrm{y}}^1=-4\mathrm{y} \end{gathered}$$