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If $y=\frac{\log x}{x}$, then the value of $x^2 \frac{d^2 y}{d x^2}+3 x \frac{d y}{d x}+y$ at the point $(\sqrt[3]{e}, \sqrt{e})$ is
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$\begin{aligned} & \text { } y=\frac{\log x}{x} \\ & \Rightarrow \frac{d y}{d x}=\frac{1-\log x}{x^2} \\ & \text { and } \frac{d^2 y}{d x^2}=\frac{-3 x+2 x \log x}{x^4} \\ & \text { Now Consider, } \\ & x^2 \frac{d^2 y}{d x^2}+3 x \frac{d y}{d x}+y \\ & =x^2\left(\frac{-3 x+2 x \log x}{x^4}\right)+3 x\left(\frac{1-\log x}{x^2}\right)+\frac{\log x}{x}=0 \forall x, y\end{aligned}$
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