For the differential equation $\left(x \log _{e} x\right) d y=\left(\log _{e} x-y\right) d x$

(A) Degree of the given differential equation is $1$.

(B) It is a homogeneous differential equation.

(C) Solution is $2y \log _{\mathrm{e}} \mathrm{x}+A=\left(\log _{\mathrm{e}} \mathrm{x}\right)^{2}$, where $A$ is an arbitrary constant

(D) Solution is $2 y \log _{e} x+A=\log _{e}\left(\log _{e} x\right)$, where $A$ is an arbitrary constant

Choose the correct answer from the options given below :

Option (A): Degree of the given differential equation is 1.

Rearranging the equation to standard form:

$\dfrac{dy}{dx} = \dfrac{\log_e x - y}{x \log_e x}$

Since $\dfrac{dy}{dx}$ has a power of 1, the degree is 1.

Option (A) is correct.

---

Option (B): It is a homogeneous differential equation.

A differential equation is homogeneous if $f(tx,ty) = f(x,y)$ for any non-zero t.

For our equation: $f(x,y) = \dfrac{\log_e x - y}{x \log_e x}$

Substituting $(tx,ty)$:

$f(tx,ty) = \dfrac{\log_e(tx) - ty}{tx \log_e(tx)} = \dfrac{\log_e t + \log_e x - ty}{tx(\log_e t + \log_e x)}$

This is not equal to $f(x,y)$ due to the $\log_e t$ terms.

Option (B) is incorrect.

---

Options (C) & (D): Solving the equation

Rearranging to group terms:

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

Let's define $u = \log_e x$, which means $du = \dfrac{dx}{x}$.

Substituting, our equation becomes:

$u\,dy + y\,du = u\,du$

This is the differential of the product $uy$:

$d(uy) = u\,dy + y\,du = u\,du$

Integrating both sides:

$uy = \dfrac{u^2}{2} + C$

Substituting back $u = \log_e x$:

$y \log_e x = \dfrac{(\log_e x)^2}{2} + C$

Multiplying by 2:

$2y \log_e x = (\log_e x)^2 + 2C$

Let $A = 2C$, so:

$2y \log_e x + A = (\log_e x)^2$

Option (C) is correct, and Option (D) is incorrect.

---

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