Question

For the differential equation \((x \log x) \, dy = (\log x - y) \, dx\):
(A) Degree of the given differential equation is 1.
(B) It is a homogeneous differential equation.
(C) Solution is \(2y \log x + A = (\log x)^2\), where \(A\) is an arbitrary constant.
(D) Solution is \(2y \log x + A = \llog(\ln x)\), where \(A\) is an arbitrary constant.

• A. (A) and (C) only
• B. (A), (B), and (C) only
• C. (A), (B), and (D) only
• D. (A) and (D) only

Hint:

When analyzing first-order linear differential equations, carefully identify the degree and whether the equation is homogeneous. For solving, try rearranging the equation and use the method of integrating factors or other appropriate techniques to find the solution. It is essential to verify the solution by substituting it back into the original equation. In this case, simplifying and integrating the equation gives us the correct solution.

Answer 1

The Correct Option is A

The given differential equation is:

\((x \log_e x) \, dy = (\log_e x - y) \, dx.\)

Rearranging:

\((x \log_e x) \, dy + y \, dx = \log_e x \, dx.\)

This is a first-order linear differential equation.

(A) Degree of the given differential equation is 1. This is correct because the highest power of the derivatives is 1.

(B) It is a homogeneous differential equation. This is incorrect because it does not meet the condition for a homogeneous equation. A homogeneous differential equation must be of the form where both sides are a function of \( \frac{dy}{dx} \), and this equation does not fit that form.

(C) Solution is \( 2y \log_e x + A = (\log_e x)^2 \), where \( A \) is an arbitrary constant. This is the correct solution to the differential equation. To solve the equation, integrate and simplify to get the solution:

\(\int \frac{dy}{dx} = \log_e x \implies 2y \log_e x = (\log_e x)^2 + A.\)

Thus, (C) is correct.

(D) Solution is \( 2y \log_e x + A = \log_e (\log_e x) \), where \( A \) is an arbitrary constant. This is incorrect as the solution derived from the equation does not match this form.

Thus, the correct answer is: (A) - (C) only.

Answer 2

The given differential equation is: 

\((x \log_e x) \, dy = (\log_e x - y) \, dx.\)

Rearranging the equation:

\[ (x \log_e x) \, dy + y \, dx = \log_e x \, dx. \]

This is a first-order linear differential equation:

The equation can be written as: \[ (x \log_e x) \, dy + y \, dx = \log_e x \, dx. \]

(A) Degree of the given differential equation:

The degree of a differential equation is the highest power of the derivative. Here, the highest power of the derivative is 1, so the degree is 1. Therefore, option (A) is correct.

(B) It is a homogeneous differential equation:

A differential equation is homogeneous if it can be written in the form where both sides involve \( \frac{dy}{dx} \) and the terms are of the same degree. However, this equation does not satisfy the condition for homogeneity. Therefore, option (B) is incorrect.

(C) Solution is \( 2y \log_e x + A = (\log_e x)^2 \):

To solve the equation, we first rearrange it as: \[ \frac{dy}{dx} = \frac{\log_e x}{x \log_e x} - \frac{y}{x \log_e x}. \] After integration and simplification, we get the solution: \[ 2y \log_e x = (\log_e x)^2 + A. \] This confirms that option (C) is correct.

(D) Solution is \( 2y \log_e x + A = \log_e (\log_e x) \):

This is incorrect because the solution derived from the differential equation does not match this form. Therefore, option (D) is incorrect.

Thus, the correct answer is:

(A) - (C) only.