The integrating factor of the differential equation \(x\displaystyle\frac{dy}{dx} + (x + 1)y = e^{-x}\), \(x \gt 0\), is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(xe^{x}\)
Dividing by x puts the equation in the linear form \(\frac{dy}{dx} + Py = Q\):
\( \displaystyle \frac{dy}{dx} + \left(1 + \frac{1}{x}\right)y = \frac{e^{-x}}{x} \)
So \(P = 1 + \frac{1}{x}\), and for \(x \gt 0\)
\( \displaystyle \int \left(1 + \frac{1}{x}\right)dx = x + \log x \)
The integrating factor is
\( e^{x + \log x} = e^{x}\cdot e^{\log x} = xe^{x} \)
Why the other options are incorrect
Option (1) \(e^{x} + x\) — Incorrect. The exponential of a sum is a product: \(e^{x + \log x} = e^{x}\cdot x\), not \(e^{x} + x\).
Option (3) \(\frac{e^{x}}{x}\) — Incorrect. This would need \(P = 1 - \frac{1}{x}\); here the term \(\frac{1}{x}\) has a plus sign.
Option (4) \(e^{x}\log x\) — Incorrect. \(e^{\log x}\) is x, not \(\log x\).