Question

An integrating factor of the differential equation $x\frac{dy}{dx}+y\log x=x{e}^x{x}^{-\frac{1}{2}\log x},(x>0)$

• (A) $x^{\log x}$
• (B) $(\sqrt{x}{)}^{\log x}$
• (C) $\{\sqrt{e}{\}}^{\{\log x{\}}^2}$
• (D) $e^{x^2}$

Answer 1

The Correct Option is C

Explanation:
Given differential equation is$$\begin{array}{l}x\frac{dy}{dx}+y\log x=x{e}^x{x}^{-1/2\log x}\\ \frac{dy}{dx}+y\frac{1}{x}\log x=e^x{x}^{-1/2\log x}\\ \text{ Here, }P=\frac{1}{x}\log x\text{ and }Q=e^x{x}^{-1/2\log x}\\ \therefore IF=e^{\int \frac{1}{x}\log xdx}\end{array}$$$$\begin{array}{l}=e^{\frac{(\log x{)}^2}{2}}\\ {=(\sqrt{e}{)}^{\log x})}^2\end{array}$$