Question

The integrating factor of the differential equation $3x{\log }_{e}x\frac{dy}{dx}+y=2{\log }_{e}x$ is given by:

• A. $({\log }_{e}x{)}^2$
• B. ${\log }_e\left({\log }_{e}x\right)$
• C. ${\log }_{e}x$
• D. $({\log }_{e}x{)}^{\frac{1}{3}}$

Answer 1

The correct option is D $({\log }_{e}x{)}^{\frac{1}{3}}$

Given,

$3x{\log }_{e}x\frac{dy}{dx}+y=2{\log }_{e}x$

Dividing both sides by $3x{\log }_{e}x$, we get
$\frac{dy}{dx}+\frac{1}{3x{\log }_{e}x}y=\frac{2{\log }_{e}x}{3x{\log }_{e}x}$

$\Rightarrow \frac{dy}{dx}+\frac{1}{3x{\log }_{e}x}y=\frac{2}{3x}$

which is linear form $\frac{dy}{dx}+Py=Q$,

where $P$ and $Q$ are function of $x$ and the integrating factor is given by following formula $e^{\int Pdx}$.

$\therefore IF=e^{\int \frac{1}{3x{\log }_{e}x}}dx$

Put ${\log }_{e}x=t\Rightarrow \frac{1}{x}dx=dt$
$=e^{\frac{1}{3}}\int \frac{dt}{t}=e^{\frac{1}{3}}\log t$

$=e^{\log t^{\frac{1}{3}}}=t^{\frac{1}{3}}=({\log }_{e}x{)}^{\frac{1}{3}}$