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

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

Taking logarithms of both sides:

    \(y = x\log_e x\)

Differentiating twice:

    \( \displaystyle \frac{dy}{dx} = \log_e x + 1 \)

    \( \displaystyle \frac{d^2y}{dx^2} = \frac{1}{x} \)

Then

    \( \displaystyle y\frac{d^2y}{dx^2} = x\log_e x \times \frac{1}{x} = \log_e x = \frac{dy}{dx} - 1 \)

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