The integrating factor of the differential equation \(\displaystyle\frac{dy}{dx} + y = \frac{1 + y}{x}\) is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{e^x}{x}\)
The right side also contains y, so the equation must first be put in the linear form \(\frac{dy}{dx} + Py = Q\).
\( \displaystyle \frac{dy}{dx} + y = \frac{1}{x} + \frac{y}{x} \)
\( \displaystyle \frac{dy}{dx} + \left(1 - \frac{1}{x}\right)y = \frac{1}{x} \)
So
\( \displaystyle P = 1 - \frac{1}{x} \)
\( \displaystyle \int P\,dx = x - \log x \)
The integrating factor is
\( \text{I.F.} = e^{\int P\,dx} = e^{x - \log x} \)
\( \displaystyle = e^x \cdot e^{-\log x} = \frac{e^x}{x} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{x}{e^x}\) — Incorrect. This is \(e^{-(x - \log x)}\), the reciprocal of the integrating factor.
Option (3) \(xe^x\) — Incorrect. This is \(e^{x + \log x}\), which needs P = 1 + 1/x; the term y/x moved to the left side gives −1/x.
Option (4) \(e^x\) — Incorrect. This takes P = 1 from the left side only and ignores the y/x on the right side.