Question

The integrating factor of the differential equation $3x{y}^{\prime }-y=1+\log x,x>0$ is

• A. $\log x$
• B. $\frac{1}{x}$
• C. $x^{-1/3}$
• D. $\frac{1}{x^3}$
• E. $x^{1/3}$

Answer 1

Answer: (C)

$y^{\prime }-\frac{1}{3x}y=\frac{1+\log x}{3x}$
$I\cdot F=e^{\int -\frac{1}{3x}dx}=e^{-\frac{1}{3}\log x}$
$=e^{\log x^{\frac{-1}{3}}}$
$=x^{-1/3}$