Question

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

• 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 $(\sqrt{e}{)}^{(\log x{)}^2}$

Given differential equation is
$x\frac{dy}{dx}+y\log x=x{e}^x{x}^{(-1/2)\log x}$
or $\frac{dy}{dx}+y\cdot \frac{1}{x}\log x=e^x{x}^{-\left(1/2\right)\log x}$
Here, $P=\frac{1}{x}\log x$ and $Q=e^x{x}^{-\left(1/2\right)\log x}$
$\therefore IF=e^{\int \frac{1}{x}\log xdx}$

Put $\log x=t$

$\frac{1}{x}dx=dt$

$IF=e^{\int tdt}$

$IF=e^{\frac{t^2}{2}}$

$=e^{\frac{(\log x{)}^2}{2}}=(\sqrt{e}{)}^{(\log x{)}^2}$.