Analyzing and Solving the Given Differential Equation

The given differential equation is \( (x \log_e x) dy = (\log_e x - y) dx \).

Let's first rewrite the equation in the standard form \(\frac{dy}{dx} = f(x, y)\) or a linear form \(\frac{dy}{dx} + P(x)y = Q(x)\) to better understand its properties.

Dividing both sides by \( (x \log_e x) dx \), we get:

\[ \frac{dy}{dx} = \frac{\log_e x - y}{x \log_e x} \] \[ \frac{dy}{dx} = \frac{\log_e x}{x \log_e x} - \frac{y}{x \log_e x} \] \[ \frac{dy}{dx} = \frac{1}{x} - \frac{y}{x \log_e x} \] Rearranging this equation to the linear form \(\frac{dy}{dx} + P(x)y = Q(x)\): \[ \frac{dy}{dx} + \left(\frac{1}{x \log_e x}\right)y = \frac{1}{x} \] This is indeed a first-order linear differential equation, where \( P(x) = \frac{1}{x \log_e x} \) and \( Q(x) = \frac{1}{x} \).

Analyzing Statement (A): Degree of the Differential Equation

The degree of a differential equation is the power of the highest order derivative when the equation is expressed in a polynomial form concerning derivatives. In the rewritten form, the highest order derivative is \(\frac{dy}{dx}\), and its power is 1. The original form can also be seen as a polynomial in \(dy/dx\) after dividing by \(dx\).

The equation involves the first derivative \( \frac{dy}{dx} \) raised to the power of 1. There are no higher-order derivatives or powers of \( \frac{dy}{dx} \) other than 1. Therefore, the degree of the given differential equation is 1.

Statement (A) is true.

Analyzing Statement (B): Homogeneous Differential Equation

A first-order differential equation \( \frac{dy}{dx} = f(x, y) \) is homogeneous if the function \( f(x, y) \) is a homogeneous function of degree zero, i.e., \( f(\lambda x, \lambda y) = f(x, y) \) for any non-zero constant \( \lambda \). In the linear form \(\frac{dy}{dx} + P(x)y = Q(x)\), it is homogeneous only if \( Q(x) = 0 \).

In our equation, \( \frac{dy}{dx} = \frac{1}{x} - \frac{y}{x \log_e x} \). Let's test the homogeneity condition on \( f(x, y) = \frac{1}{x} - \frac{y}{x \log_e x} \).

\[ f(\lambda x, \lambda y) = \frac{1}{\lambda x} - \frac{\lambda y}{(\lambda x) \log_e (\lambda x)} = \frac{1}{\lambda x} - \frac{\lambda y}{\lambda x (\log_e \lambda + \log_e x)} = \frac{1}{\lambda x} - \frac{y}{x (\log_e \lambda + \log_e x)} \] This is not equal to \( f(x, y) = \frac{1}{x} - \frac{y}{x \log_e x} \).

Also, in the linear form, \( Q(x) = \frac{1}{x} \), which is not zero. Therefore, the equation is not homogeneous.

Statement (B) is false.

Solving the Differential Equation

We have the linear differential equation \(\frac{dy}{dx} + \left(\frac{1}{x \log_e x}\right)y = \frac{1}{x}\). The integrating factor (IF) is given by \( e^{\int P(x) dx} \).

Let's calculate the integral of \( P(x) \): \[ \int P(x) dx = \int \frac{1}{x \log_e x} dx \] Let \( u = \log_e x \). Then \( du = \frac{1}{x} dx \). The integral becomes: \[ \int \frac{1}{u} du = \log_e |u| \] Assuming \( \log_e x > 0 \) (which implies \( x > 1 \)), we have \( \log_e (\log_e x) \). So, the integrating factor is: \[ \text{IF} = e^{\int P(x) dx} = e^{\log_e (\log_e x)} = \log_e x \] The general solution of a first-order linear differential equation is given by \( y \times \text{IF} = \int Q(x) \times \text{IF} dx + A \), where A is the arbitrary constant.

Substitute the values of \( y, \text{IF}, \) and \( Q(x) \): \[ y (\log_e x) = \int \left(\frac{1}{x}\right) (\log_e x) dx + A \] Now, we need to evaluate the integral \( \int \frac{\log_e x}{x} dx \). Let \( v = \log_e x \). Then \( dv = \frac{1}{x} dx \). The integral becomes: \[ \int v dv = \frac{v^2}{2} + C_1 = \frac{(\log_e x)^2}{2} + C_1 \] Substitute this back into the solution equation: \[ y \log_e x = \frac{(\log_e x)^2}{2} + A \] To match the format of the options, let's rearrange this equation:

Multiply by 2:

\[ 2y \log_e x = (\log_e x)^2 + 2A \] Move the constant term to the left side (or rearrange): \[ 2y \log_e x - (\log_e x)^2 = 2A \] Let \( A' = -2A \) (or \( A' = -2A \) depending on how the constant is defined). The solution can be written as:

\[ 2y \log_e x + A' = (\log_e x)^2 \] Using A as the arbitrary constant as in the options:

\[ 2y \log_e x + A = (\log_e x)^2 \]

Analyzing Statements (C) and (D) based on the Solution

The derived solution is \( 2y \log_e x + A = (\log_e x)^2 \), where A is an arbitrary constant.

Statement (C): Solution is \( 2y \log_e x + A = (\log_e x)^2 \), where A is an arbitrary constant.

This matches our derived solution.

Statement (C) is true.

Statement (D): Solution is \( 2y \log_e x + A = \log_e (\log_e x) \), where A is an arbitrary constant.

This does not match our derived solution.

Statement (D) is false.

Conclusion based on Statements

• Statement (A) is true.
• Statement (B) is false.
• Statement (C) is true.
• Statement (D) is false.

The correct statements are (A) and (C).

Now let's check the given options:

• Option 1: (A) and (C) only - This aligns with our findings.
• Option 2: (A), (B) and (C) only - Incorrect, as (B) is false.
• Option 3: (A), (B) and (D) only - Incorrect, as (B) and (D) are false.
• Option 4: (A) and (D) only - Incorrect, as (D) is false.

Therefore, the correct answer is Option 1, stating that only (A) and (C) are correct.

Revision Table: Differential Equation Concepts

Additional Information: Types of First-Order Differential Equations

First-order differential equations can be classified into several types, each with its own method of solution:

• Variable Separable: Equations that can be written as \( g(y) dy = f(x) dx \). Integrate both sides to find the solution.
• Homogeneous: Equations of the form \(\frac{dy}{dx} = f(x,y)\) where \(f(x,y)\) is homogeneous of degree zero. Solved by substituting \( y = vx \) and transforming the equation into a variable separable form in terms of v and x.
• Equations Reducible to Homogeneous Form: Equations of the form \(\frac{dy}{dx} = \frac{a_1x + b_1y + c_1}{a_2x + b_2y + c_2}\). The method of solution depends on whether the lines \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) are intersecting or parallel.
• Linear: Equations of the form \(\frac{dy}{dx} + P(x)y = Q(x)\). Solved using an integrating factor.
• Exact Differential Equations: Equations of the form \( M(x, y) dx + N(x, y) dy = 0 \) where \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). Solved by finding a function \( \phi(x, y) \) such that \( d\phi = M dx + N dy \).
• Equations Reducible to Exact Form: Equations that are not exact but can be made exact by multiplying by a suitable integrating factor.
• Bernoulli's Equation: Equations of the form \(\frac{dy}{dx} + P(x)y = Q(x)y^n\). This can be reduced to a linear equation by substituting \( v = y^{1-n} \).

The equation in this problem is a first-order linear differential equation, which is a common and important type.