A curve has equation ${y={{x}^{{{x}^{-1}}}}}$, ${x\in \mathbb{R}}$. Show that ${y}$ is a solution of

Question

A curve has equation ${y={{x}^{{{x}^{-1}}}}}$, ${x\in \mathbb{R}}$. Show that ${y}$ is a solution of the following differential equation. ${{{x}^{3}}\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}-x\left( 1-\ln x \right)\frac{\mathrm{d}y}{\mathrm{d}x}+\left( 3-2\ln x \right)y=0}$. Show further that ${{{\left. \frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}} \right|}_{x=\mathrm{e}}}=-{{\mathrm{e}}^{{{\mathrm{e}}^{-1}}-3}}}$.

Accepted Answer

$\frac{d^2y}{dx^2}\bigg|_{x=e} = -e^{e^{-1}-3}$

Techniques and Applications Of Differentiation Q1

Techniques and Applications Of Differentiation Q1

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