Question

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

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

Answer 1

The correct option is B $(\sqrt{x}{)}^{\log x}$

$\frac{dy}{dx}+\left(\frac{\log x}{x}\right)y=e^x{x}^{-\frac{1}{2}\log x}$
I.F. $=\exp \left\{\int \frac{\log x}{x}dx\right\}$
$=e^{\frac{1}{2}(\log x{)}^2}={\left(e^{\frac{1}{2}\left(\log x\right)}\right)}^{\log x}$
$={\left(e^{\log \sqrt{x}}\right)}^{\log x}=(\sqrt{x}{)}^{\log x}$