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  4. The integrating factor of the differential equation 3 x y'-y=1+ log x, x > 0 is

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Q. The integrating factor of the differential equation $3 x y'-y=1+\log x, x > 0$ is

KEAMKEAM 2020

Solution:

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