For the differential equation ≤ft(xlog_e x)dy=≤ft(log_e x-y)dx (A) De…

The correct answer is Option (1) → (A) and (C) only

Dividing by \(x\log_e x\):

    \( \displaystyle \frac{dy}{dx} = \frac{\log_e x - y}{x\log_e x} \)

    \( \displaystyle \frac{dy}{dx} + \frac{1}{x\log_e x}\,y = \frac{1}{x} \)

This is a linear differential equation with \(P = \frac{1}{x\log_e x}\) and \(Q = \frac{1}{x}\). Since \(\frac{d}{dx}\left(\log_e(\log_e x)\right) = \frac{1}{x\log_e x}\), the integrating factor is

    \( \displaystyle \text{I.F.} = e^{\int\frac{dx}{x\log_e x}} = e^{\log_e(\log_e x)} = \log_e x \)

The solution is \(y \times \text{I.F.} = \int Q \times \text{I.F.}\,dx\):

    \( \displaystyle y\log_e x = \int\frac{\log_e x}{x}\,dx \)

    \( \displaystyle y\log_e x = \frac{(\log_e x)^2}{2} + C \)

    \(2y\log_e x - 2C = (\log_e x)^2\)

So the equation has order 1 and degree 1, its right side \(\frac{\log_e x - y}{x\log_e x}\) is not a function of \(\frac{y}{x}\) alone, and its solution is \(2y\log_e x + A = (\log_e x)^2\) with A = −2C.

Checking the statements

(A) Degree of the given differential equation is 1. — Correct. The only derivative, dy/dx, occurs to the power 1.

(B) It is a homogeneous differential equation. — Incorrect. Replacing x, y by λx, λy turns \(\log_e x\) into \(\log_e x + \log_e\lambda\), so the right side is not of degree zero.

(C) Solution is \(2y\log_e x+A=\left(\log_e x\right)^2\), where A is an arbitrary constant — Correct. It is the solution above with A = −2C.

(D) Solution is \(2y\log_e x+A=\log_e\left(\log_e x\right)\), where A is an arbitrary constant — Incorrect. The right side of the solution is \((\log_e x)^2\), not \(\log_e(\log_e x)\).