(a)OR(b)

If $\displaystyle \mathrm{y}=\log \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}$, then show that $\displaystyle x(x+1)^{2} \mathrm{y}_{2}+(x+1)^{2} \mathrm{y}_{1}=2$.

The given function can be written as \[\begin{aligned} & \mathrm{y}=2 \log (x+1)-\log x \\ & \Rightarrow \mathrm{y}_{1}=\frac{2}{x+1}-\frac{1}{x}=\frac{x-1}{x(x+1)} \\ & \Rightarrow(x+1) \mathrm{y}_{1}=\frac{x-1}{x}=1-\frac{1}{x} \end{aligned} \] \[\begin{aligned} & x \sqrt{1+\mathrm{y}}+\mathrm{y} \sqrt{1+x}=0 \\ & \Rightarrow x \sqrt{1+\mathrm{y}}=-\mathrm{y} \sqrt{1+x} \\ & \Rightarrow x^{2}(1+\mathrm{y})=\mathrm{y}^{2}(1+x) \\ & \Rightarrow(x-\mathrm{y})(x+\mathrm{y})+x \mathrm{y}(x-\mathrm{y})=0 \\ & \Rightarrow(x-\mathrm{y})(x+\mathrm{y}+x \mathrm{y})=0 \end{aligned} \] $\displaystyle x \neq \mathrm{y} \Rightarrow x+\mathrm{y}+x \mathrm{y}=0$ \[\Rightarrow \mathrm{y}=\frac{-x}{1+x} \] \[\Rightarrow \frac{d \mathrm{y}}{d x}=\frac{-1}{(1+x)^{2}} \]