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

Answer: (A)

(A) and (C) only

Answer 2

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.