(a)OR(b)

If $\displaystyle \mathrm{y}=(\log \mathrm{x})^{2}$, prove that $\displaystyle \mathrm{x}^{2} \mathrm{y}^{\prime \prime}+\mathrm{xy}^{\prime}=2$.

Differentiating both sides w.r.t. x , we get $$\begin{aligned} & \mathrm{y}^{\prime}=\frac{2 \log \mathrm{x}}{\mathrm{x}} \\ & \Rightarrow \mathrm{x} \mathrm{y}^{\prime}=2 \log \mathrm{x} \\ & \Rightarrow \mathrm{x} \mathrm{y}^{\prime \prime}+\mathrm{y}^{\prime}=\frac{2}{\mathrm{x}} \\ & \Rightarrow \mathrm{x}^{2} \mathrm{y}^{\prime \prime}+\mathrm{xy}^{\prime}=2 \end{aligned} \begin{aligned} & \frac{d \mathrm{y}}{d \mathrm{x}}=\cos \left(\tan ^{-1}\left(\mathrm{e}^{\mathrm{x}}\right)\right) \times \frac{\mathrm{e}^{\mathrm{x}}}{1+\mathrm{e}^{2 \mathrm{x}}} \\ & \left(\frac{d \mathrm{y}}{d \mathrm{x}}\right)_{\mathrm{x}=0}=\cos \frac{\pi}{4} \times \frac{1}{2}=\frac{1}{2 \sqrt{2}} \end{aligned} $$