For the differential equation (xlog_e x)dy = (log_ex − y)dx:

(A) Degree of the given differential equation is 1.
(B) It is a homogeneous differential equation.
(C) Solution is 2yloge x + A = (logex)2, where A is an arbitrary constant
(D) Solution is 2yloge x + A = loge(logex), where A is an arbitrary constant

Choose the correct answer from the options given below:

The correct answer is (A) and (C) only.

Let us analyze the given differential equation step-by-step to understand why statements (A) and (C) are correct, while statement (B) is incorrect.

1. Analysis of Statement (A): Degree of the differential equation
The given differential equation is:
$$\left(x{log}_{e}x\right)dy=({log}_{e}x-y)dx$$
We can rewrite this by dividing both sides by $dx$:
$$x{log}_{e}x\frac{dy}{dx}={log}_{e}x-y$$
The highest order derivative present in this equation is $\frac{dy}{dx}$, which is of first order. Since the differential equation is a polynomial in $\frac{dy}{dx}$, and the power of this highest order derivative is 1, the degree of the differential equation is 1. Therefore, statement (A) is correct.

2. Analysis of Statement (B): Homogeneity of the differential equation
We can express the derivative $\frac{dy}{dx}$ as:
$$\frac{dy}{dx}=\frac{{log}_{e}x-y}{x{log}_{e}x}$$
A differential equation of the form $\frac{dy}{dx}=f(x,y)$ is homogeneous if $f(\lambda x,\lambda y)=f(x,y)$ for any non-zero constant $\lambda$. Here, replacing $x$ with $\lambda x$ and $y$ with $\lambda y$ yields terms containing ${log}_e\left(\lambda x\right)={log}_e\lambda +{log}_{e}x$, which prevents the expression from simplifying back to $f(x,y)$. Thus, the differential equation is not homogeneous, and statement (B) is incorrect.

3. Analysis of Statements (C) and (D): Solving the differential equation
Let us solve the differential equation. Rearranging the terms to form a standard linear differential equation:
$$x{log}_{e}x\frac{dy}{dx}+y={log}_{e}x$$
Dividing the entire equation by $x{log}_{e}x$:
$$\frac{dy}{dx}+\frac{1}{x{log}_{e}x}y=\frac{1}{x}$$
This is a first-order linear differential equation of the form $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$, where:
$$P\left(x\right)=\frac{1}{x{log}_{e}x}\text{ and }Q\left(x\right)=\frac{1}{x}$$
First, we find the integrating factor (I.F.):
$$\text{I.F.}=e^{\int P\left(x\right)dx}=e^{\int \frac{1}{x{log}_{e}x}dx}$$
To evaluate the integral, let $u={log}_{e}x$, which gives $du=\frac{1}{x}dx$:
$$\int \frac{1}{x{log}_{e}x}dx=\int \frac{1}{u}du={log}_{e}u={log}_e\left({log}_{e}x\right)$$
Substituting this back into the integrating factor:
$$\text{I.F.}=e^{{log}_e\left({log}_{e}x\right)}={log}_{e}x$$
The general solution of the linear differential equation is given by:
$$y·\left(\text{I.F.}\right)=\int Q\left(x\right)·\left(\text{I.F.}\right)dx+C$$
Substituting the values of $\text{I.F.}$ and $Q\left(x\right)$:
$$y{log}_{e}x=\int \frac{1}{x}{log}_{e}xdx+C$$
Using the substitution $u={log}_{e}x$ and $du=\frac{1}{x}dx$ again:
$$\int \frac{1}{x}{log}_{e}xdx=\int udu=\frac{u^2}{2}=\frac{{\left({log}_{e}x\right)}^2}{2}$$
So, the equation becomes:
$$y{log}_{e}x=\frac{{\left({log}_{e}x\right)}^2}{2}+C$$
Multiplying the entire equation by 2:
$$2y{log}_{e}x={\left({log}_{e}x\right)}^2+2C$$
Letting $-2C=A$ (where $A$ is an arbitrary constant):
$$2y{log}_{e}x+A={\left({log}_{e}x\right)}^2$$
This matches the expression in statement (C). Thus, statement (C) is correct, and statement (D) is incorrect.

Since statements (A) and (C) are correct, the correct option is (A) and (C) only.