Question

If $y=x^x$, prove that $\frac{d^{2}y}{d{x}^2}-\frac{1}{y}{\left(\frac{dy}{dx}\right)}^2-\frac{y}{x}=0.$

Answer 1

$y=x^x$

$logy=xlogx$

$\frac{1}{y}\frac{dy}{dx}=1+logx$

$\frac{dy}{dx}=y(1+logx)=y^{\prime }$--- (1)

$\frac{d^{2}y}{d{x}^2}=y\times \frac{1}{x}+(1+logx)\frac{dy}{dx}$

$\frac{d^{2}y}{d{x}^2}=\frac{y}{x}+\frac{y^{\prime }}{y}.y^{\prime }$

$\frac{d^{2}y}{d{x}^2}-\frac{1}{y}(\frac{dy}{dx}{)}^2-\frac{y}{x}=0$

Hence proved.