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

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.