The integrating factor of the differential equation :
\(\frac{dy}{dx}\) (x log x) + y = 2 log x
is :
Answer & explanation
Correct answer: option 2
\(\frac{dy}{dx}\) (x log x) + y = 2 log x
\(\frac{dy}{dx}\) + \(\frac{y}{(x log x)}\) = \(\frac{2}{x}\)
Compare with : y' + yP(x) = Q(x)
Integrating factor : e\(\int_{}^{}d\)x P= e \(\int_{}^{}d\)x 1/x log x = e log(log x) = log x