For the differential equation \(x\displaystyle\frac{dy}{dx}+2y= x^2\log_ex\)
(A) Integrating factor is \(2x\)
(B) Integrating factor is \(x^2\)
(C) General Solution is \(y=\displaystyle\frac{x^2}{16}(4 \log_e|x|-1)+Cx^{-2}\) where C is an arbitrary constant.
(D) General Solution is \(y=\displaystyle\frac{x^2}{16}(4 \log_e|x|-1)+C\) where C is an arbitrary constant.
Choose the correct answer from the options given below.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B) and (C) only
Since \(\log_e x\) appears, x > 0, so \(|x| = x\). Dividing by x gives a linear differential equation \(\frac{dy}{dx} + Py = Q\):
\( \displaystyle \frac{dy}{dx} + \frac{2}{x}\,y = x\log_e x \)
\( \displaystyle \text{I.F.} = e^{\int \frac{2}{x}\,dx} = e^{2\log_e x} = x^2 \)
The solution is \(y \cdot \text{I.F.} = \int Q \cdot \text{I.F.}\,dx\):
\( \displaystyle x^2y = \int x^3\log_e x\,dx \)
By integration by parts, with \(\log_e x\) as the first function:
\( \displaystyle \int x^3\log_e x\,dx = \frac{x^4}{4}\log_e x - \int \frac{x^4}{4}\cdot\frac{1}{x}\,dx \)
\( \displaystyle = \frac{x^4}{4}\log_e x - \frac{x^4}{16} + C \)
\( \displaystyle x^2y = \frac{x^4}{16}(4\log_e x - 1) + C \)
Dividing by \(x^2\):
\( \displaystyle y = \frac{x^2}{16}(4\log_e x - 1) + Cx^{-2} \)
So the integrating factor is \(x^2\), and the general solution is \(y = \frac{x^2}{16}(4\log_e|x| - 1) + Cx^{-2}\).
Checking the statements
(A) Integrating factor is \(2x\) — Incorrect. \(e^{\int \frac{2}{x}dx} = x^2\), not 2x.
(B) Integrating factor is \(x^2\) — Correct. \(e^{2\log_e x} = x^2\).
(C) General Solution is \(y=\displaystyle\frac{x^2}{16}(4 \log_e|x|-1)+Cx^{-2}\) where C is an arbitrary constant. — Correct. The constant is divided by \(x^2\), giving \(Cx^{-2}\).
(D) General Solution is \(y=\displaystyle\frac{x^2}{16}(4 \log_e|x|-1)+C\) where C is an arbitrary constant. — Incorrect. The constant term must be \(Cx^{-2}\), not C.
NCERT (2026-27 edition):
Linear Differential Equations
