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

Question

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

Dr. Abubacker · Accepted Answer

(c)
Given, $\frac{dy}{dx} + y.\frac{1}{x}\log x = e^{x}\ x^{- \left( \frac{1}{2} ight)\log x}$
$	herefore$ $IF = e^{\int_{}^{}{\frac{1}{x}\log x\ dx}} = e^{\int_{}^{}\frac{\left( \log x ight)^{2}}{2}} = \left( \sqrt{e} ight)^{\left( \log x ight)^{2}}$

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

Explanation

Reviewed by Dr. Abubacker

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