Question

The integrating factor of $x \cdot \frac{dy}{dx} + y \log x = x \cdot e^x x^{-1/2} \log x$ is

• A. $(\log x)^x$
• B. $x^{\log x}$
• C. $(\sqrt{x})^{\log x}$
• D. $e^{\sqrt{x} \log x}$

Hint:

$\int \frac{\log x}{x} dx = \frac{(\log x)^2}{2}$.

Answer 1

The Correct Option is C

Step 1: Standard Form
Divide by $x$: $\frac{dy}{dx} + \frac{\log x}{x} y = \dots$.
Here, $P = \frac{\log x}{x}$.

Step 2: Calculate $I.F.$
$I.F. = e^{\int P dx} = e^{\int \frac{\log x}{x} dx}$.
Let $u = \log x, du = \frac{1}{x} dx \implies \int u du = \frac{u^2}{2} = \frac{(\log x)^2}{2}$.
$I.F. = e^{\frac{1}{2}(\log x)^2}$.

Step 3: Simplify
$e^{\frac{1}{2}(\log x)^2} = (e^{\log x})^{\frac{1}{2}\log x} = x^{\frac{1}{2}\log x} = (x^{1/2})^{\log x} = (\sqrt{x})^{\log x}$.
Final Answer: (C)