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Integerating factor of (xlog⁡ex)dydx+y=2log⁡ex(x \log_e x) \frac{dy}{dx} + y = 2 \log_e x is :

Solution

✅ Correct Option: 3

Rewrite in standard form: dydx+1xlog⁡xy=2x\frac{dy}{dx}+\frac{1}{x\log x}y=\frac{2}{x}.

Integrating factor =e∫dxxlog⁡x=elog⁡(log⁡x)=log⁡x=e^{\int\frac{dx}{x\log x}}=e^{\log(\log x)}=\log x.

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