An integrating factor of the differential equation \(\displaystyle\frac{dy}{dx}+\frac{y}{x}=\frac{1}{x^2};(x \gt 0)\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(x\)
The equation is linear, of the form \(\frac{dy}{dx} + Py = Q\), with
\(\displaystyle P = \frac{1}{x}\)
The integrating factor is
\(\displaystyle \text{I.F.} = e^{\int P\,dx} = e^{\int \frac{1}{x}\,dx} = e^{\log x} = x\quad (x \gt 0)\)