If $\mathrm{y} = \frac{{\sinh }^{-1}\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}$ then $(1+{\mathrm{x}}^2){\mathrm{y}}_2 + 3{\mathrm{xy}}_1 + \mathrm{y} =$

$$\begin{gathered} \mathrm{y}=\frac{{\sin }^{-1}\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}} \\ \mathrm{y}\sqrt{1+{\mathrm{x}}^2}=\sin {\mathrm{h}}^{-1}\mathrm{x} \\ \left(1+{\mathrm{x}}^2\right){\mathrm{y}}^2={\left(\sin {\mathrm{h}}^{-1}\mathrm{x}\right)}^2 \end{gathered}$$

$$\begin{gathered} (1+{\mathrm{x}}^2)2\mathrm{y} {\mathrm{y}}_1+2{\mathrm{xy}}^2=\frac{2\sin {\mathrm{h}}^{-1}\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}} \\ (1+{\mathrm{x}}^2)2{\mathrm{yy}}_1+2{\mathrm{xy}}^2=2\mathrm{y} \\ (1+{\mathrm{x}}^2){\mathrm{y}}_1+\mathrm{xy}=1 \\ (1+{\mathrm{x}}^2){\mathrm{y}}_2+2{\mathrm{xy}}_1+{\mathrm{xy}}_1+\mathrm{y}=0 \\ (1+{\mathrm{x}}^2){\mathrm{y}}_2+3{\mathrm{xy}}_1+\mathrm{y}=0 \end{gathered}$$