The integrating factor of the differential equation, \(x^2\displaystyle\frac{dy}{dx}+xy = \log_e x\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(x\)
The integrating factor is read from the standard linear form \(\frac{dy}{dx} + Py = Q\), in which the coefficient of \(\frac{dy}{dx}\) is 1. Dividing the equation by \(x^2\) (here \(x \gt 0\), since \(\log_e x\) appears):
\( \displaystyle \frac{dy}{dx} + \frac{1}{x}\,y = \frac{\log_e x}{x^2} \)
So \(P = \frac{1}{x}\), and
\( \displaystyle \text{I.F.} = e^{\int P\,dx} = e^{\int \frac{1}{x}\,dx} \)
\( = e^{\log_e x} = x \)
Tip: Always divide by the coefficient of \(\frac{dy}{dx}\) first. Taking P = x, the coefficient of y in the given equation, gives \(e^{\int x\,dx} = e^{\frac{x^2}{2}}\), which is Option (2).