If e^y=x^x, then which of the following is true?

The correct answer is Option (4) → $y \frac{d^2 y}{d x^2}-\frac{d y}{d x}+1=0$

Given $e^y=x^x$

Taking logarithm

$y=x\log_e x$

Differentiate

$\frac{dy}{dx}=\log_e x+1$

Differentiate again

$\frac{d^2y}{dx^2}=\frac{1}{x}$

Substitute in $y\frac{d^2y}{dx^2}-\frac{dy}{dx}+1$

$=x\log_e x\cdot\frac{1}{x}-(\log_e x+1)+1$

$=\log_e x-\log_e x-1+1=0$

The correct relation is $y\frac{d^2y}{dx^2}-\frac{dy}{dx}+1=0$.