For the differential equation \(x\displaystyle\frac{dy}{dx}+3y= x^2\log_e x\), which of the following statements are TRUE?
(A) Product of order and degree is 1
(B) Integrating factor is \(x^3\)
(C) Integrating factor is \(3x\)
(D) General solution is \(y =\displaystyle\frac{x^3}{36} (6\log_e|x|-1)+ Cx^{-3}\), C is an arbitrary constant.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (B) only
The highest derivative is \(\frac{dy}{dx}\), to the power 1, so the order is 1 and the degree is 1. Dividing by x (x > 0, since \(\log_e x\) appears) gives a linear differential equation:
\(\displaystyle \frac{dy}{dx} + \frac{3}{x}\,y = x\log_e x\)
The integrating factor is
\(\displaystyle \text{I.F.} = e^{\int \frac{3}{x}\,dx} = e^{3\log_e x} = x^3\)
Multiplying by \(x^3\):
\(\displaystyle x^3y = \int x^4\log_e x\,dx\)
By integration by parts, with \(\log_e x\) as the first function:
\(\displaystyle \int x^4\log_e x\,dx = \frac{x^5}{5}\log_e x - \int \frac{x^5}{5}\cdot\frac{1}{x}\,dx\)
\(\displaystyle = \frac{x^5}{5}\log_e x - \frac{x^5}{25} + C\)
Dividing by \(x^3\):
\(\displaystyle y = \frac{x^2}{25}(5\log_e x - 1) + Cx^{-3}\)
So the order and degree are both 1, the integrating factor is \(x^3\), and the general solution is \(y = \frac{x^2}{25}(5\log_e x - 1) + Cx^{-3}\).
Checking the statements
(A) Product of order and degree is 1 — Correct. \(1 \times 1 = 1\).
(B) Integrating factor is \(x^3\) — Correct. \(e^{\int \frac{3}{x}dx} = x^3\).
(C) Integrating factor is \(3x\) — Incorrect. The integrating factor is \(x^3\), not \(3x\).
(D) General solution is \(y =\frac{x^3}{36} (6\log_e|x|-1)+ Cx^{-3}\), C is an arbitrary constant. — Incorrect. The particular part is \(\frac{x^2}{25}(5\log_e x - 1)\), not \(\frac{x^3}{36}(6\log_e|x| - 1)\).