Question

The integrating factor of the differential equation $\frac{dy}{dx}\left(x{\log }_{e}x\right)+y=2{\log }_{e}x$ is given by

• A. $x$
• B. $e^x$
• C. ${\log }_{e}x$
• D. ${\log }_e\left({\log }_{e}x\right)$

Answer 1

Answer: (A), (C)

${\log }_{e}x$
$x$

$\frac{dy}{dx}\left(x{\log }_{e}x\right)+y=2{\log }_{e}x$

$\Rightarrow \frac{dy}{dx}+\frac{1}{\left(x{\log }_{e}x\right)}y=\frac{2{\log }_{e}x}{\left(x{\log }_{e}x\right)}$

$\Rightarrow \frac{dy}{dx}+\frac{1}{\left(x{\log }_{e}x\right)}y=\frac{2}{x}$ which is of the form $\frac{dy}{dx}+py=q$

Integrating factor$=e^{\int pdx}$ where $p=\frac{1}{\left(x{\log }_{e}x\right)}$

Integrating factor$=e^{\int \frac{1}{\left(x{\log }_{e}x\right)}dx}$

Take $t={\log }_{e}x\Rightarrow dt=\frac{1}{x}dx$

$\int \frac{1}{\left(x{\log }_{e}x\right)}dx=\int \frac{dt}{t}=\ln \left|t\right|wheret={\log }_{e}x$

Integrating factor$=e^{\int pdx}=e^{\ln t}=t={\log }_{e}x$