Integrating factor of the differential equation \( x\log x\displaystyle\frac{dy}{dx}+y=2\log x \) is equal to :
(Consider \( \log x=\log_e x \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \log|x| \)
Dividing by \( x\log x \) gives the standard linear form \( \frac{dy}{dx}+Py=Q \):
\( \displaystyle \frac{dy}{dx}+\frac{1}{x\log x}\,y=\frac{2}{x} \)
\( \displaystyle P=\frac{1}{x\log x} \)
Putting \( t=\log x \), \( dt=\frac{dx}{x} \):
\( \displaystyle \int P\,dx=\int\frac{dt}{t} \)
\( \displaystyle =\log|\log x| \)
\( \displaystyle \text{IF}=e^{\int P\,dx} \)
\( \displaystyle =e^{\log|\log x|} \)
\( \displaystyle =\log x \)
Since x > 0 in \( \log x \), this is the same as \( \log|x| \).
Tip: Make the coefficient of \( \frac{dy}{dx} \) equal to 1 before reading off P; taking P = 1 from the given form gives \( e^x \), Option (3).