Question

 

If y = x^x, prove that `(d^2y)/(dx^2)−1/y(dy/dx)^2−y/x=0.`

 

Answer 1

y = x^x

Applying logarithm, 

log y = x log x

`1/y dy/dx=logx+xxx1/x=1+logx`

 

`dy/dx=x^x[1+logx]`

`(d^2y)/dx^2=(d(x^2))/dx(1+logx)+x^x[d/dx(1+logx)]`

 `=x^x(1+logx)(1+logx)+x^x[1/x]`

`=x^x(1+logx)^2+x^(x-1)`

`(d^2y)/dx^2-1/y(dy/dx)^2-y/x=x^x(1+logx)^2+x^(x-1)-1/(x^2)(x^x(1+logx)^2)-x^2/x`

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

^{= 0}

Hence proved.