For the differential equation $x\frac{dy}{dx} + 3y = x^2\log_e x$, 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 $y = \frac{x^3}{36}(6\log_e|x| - 1) + Cx^{-3}$, C is an arbitrary constant.

Choose the correct answer from the options given below:

Given differential equation: $x\frac{dy}{dx} + 3y = x^2\log_e x$

Dividing the entire equation by $x$ to get the standard linear form:

$\frac{dy}{dx} + \frac{3}{x}y = x\log_e x$

This is in the form $\frac{dy}{dx} + P(x)y = Q(x)$ where $P(x) = \frac{3}{x}$ and $Q(x) = x\log_e x$

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Checking Statement (A):

Order = highest derivative present = 1 (only $\frac{dy}{dx}$ appears)

Degree = power of the highest order derivative = 1 (power of $\frac{dy}{dx}$ is 1)

Product = Order × Degree

$= 1 × 1$

$= 1$

Statement (A) is TRUE.

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The integrating factor is given by $IF = e^{\int P(x)dx}$

$IF = e^{\int \frac{3}{x}dx}$

$= e^{3\ln|x|}$

$= e^{\ln|x|^3}$

$= x^3$

Statement (B) is TRUE.

Statement (C) is FALSE.

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Multiplying both sides by $IF = x^3$:

$x^3\frac{dy}{dx} + 3x^2y = x^4\log_e x$

The left side becomes:

$\frac{d}{dx}(x^3y) = x^4\log_e x$

Integrating both sides:

$x^3y = \int x^4\log_e x \, dx$

Using integration by parts with $u = \log_e x$, $dv = x^4dx$, then $du = \frac{1}{x}dx$, $v = \frac{x^5}{5}$:

$\int x^4\log_e x \, dx = \frac{x^5}{5}\log_e x - \int \frac{x^5}{5} \cdot \frac{1}{x}dx$

$= \frac{x^5}{5}\log_e x - \frac{x^5}{25} + C$

Therefore:

$x^3y = \frac{x^5}{5}\log_e x - \frac{x^5}{25} + C$

$y = \frac{x^2}{5}\log_e x - \frac{x^2}{25} + Cx^{-3}$

This does not match the form given in statement (D).

Statement (D) is FALSE.

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Summary:

(A) TRUE

(B) TRUE

(C) FALSE

(D) FALSE

Answer: Option 2: (A) and (B) only