Question

 

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

 

Answer 1

y = x^x

Applying logarithm, 

log y = x log x

${\displaystyle \frac{1}{y}\frac{dy}{dx}=\log x+\times x\frac{1}{x}=1+\log x}$

 

${\displaystyle \frac{dy}{dx}=x^x[1+\log x]}$

${\displaystyle \frac{d^{2}y}{{dx}^2}=\frac{d\left(x^2\right)}{dx}(1+\log x)+x^x\left[\frac{d}{dx}(1+\log x)\right]}$

 ${\displaystyle =x^x(1+\log x)(1+\log x)+x^x\left[\frac{1}{x}\right]}$

${\displaystyle =x^x{(1+\log x)}^2+x^{x-1}}$

${\displaystyle \frac{d^{2}y}{{dx}^2}-\frac{1}{y}{\left(\frac{dy}{dx}\right)}^2-\frac{y}{x}=x^x{(1+\log x)}^2+x^{x-1}-\frac{1}{x^2}\left(x^x{(1+\log x)}^2\right)-\frac{x^2}{x}}$

= x^x(1+log x)² + x^x−1 −x^x(1+log x)² − x^x−1

^{= 0}

Hence proved.