Question

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

• A. $y \frac{d^2y}{dx^2} = 1$
• B. $\frac{d^2y}{dx^2} - y = 0$
• C. $\frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$
• D. $y \frac{d^2y}{dx^2} - \frac{dy}{dx} + 1 = 0$

Answer 1

Answer: (D)

$y \frac{d^2y}{dx^2} - \frac{dy}{dx} + 1 = 0$

Answer 2

We start with the given equation:

$e^y = x^x$.

Take the natural logarithm of both sides:

$y = \ln(x^x)$.

Simplify using logarithmic properties:

$y = x \ln(x)$.

Differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \ln(x) + 1$.

Differentiate again to find the second derivative:

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

Substitute $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ into the given options. For option (4):

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

Simplify:

$\ln(x) - (\ln(x) + 1) + 1 = 0$.

Thus, option (4) satisfies the equation.