37. Consider the differential equation, $x\frac{d}{y}/\frac{d}{x}=y({\log }_{e}y-{\log }_{e}x+1)$, then which of the following are true?

(A) It is a linear differential equation
(B) It is a homogeneous differential equation
(C) Its general solution is ${\log }_e\left(\frac{y}{x}\right)=Cx$, where C is constant of integration
(D) Its general solution is ${\log }_e\left(\frac{x}{y}\right)=Cy$, where C is constant of integration
(E) If $y\left(1\right)=1$, then its particular solution is $y=x$

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Choose the correct answer from the options given below:

The correct options are (B), (C) and (E) only.

Let us analyze the given differential equation step-by-step to understand why these options are correct.
The given differential equation is:
$$x\frac{dy}{dx}=y\left({\log }_{e}y-{\log }_{e}x+1\right)$$

Using the logarithmic property ${\log }_{e}a-{\log }_{e}b={\log }_e\left(\frac{a}{b}\right)$, we can rewrite the differential equation as:
$$x\frac{dy}{dx}=y\left({\log }_e\left(\frac{y}{x}\right)+1\right)$$
Dividing both sides by $x$ gives:
$$\frac{dy}{dx}=\frac{y}{x}\left({\log }_e\left(\frac{y}{x}\right)+1\right)$$

Step 1: Check for Homogeneity (Option B)
A differential equation of the form $\frac{dy}{dx}=f(x, y)$ is homogeneous if the function $f(x, y)$ can be expressed solely as a function of the ratio $\frac{y}{x}$.
Since the right-hand side is completely in terms of $\frac{y}{x}$, we have:
$$f(x,y)=g\left(\frac{y}{x}\right)$$
Therefore, the differential equation is indeed a homogeneous differential equation. This makes Statement (B) true.

Step 2: Find the General Solution (Option C)
To solve a homogeneous differential equation, we substitute:
$$y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$$
Substituting these into our rewritten differential equation, we get:
$$v+x\frac{dv}{dx}=v\left({\log }_{e}v+1\right)$$
Expanding the right-hand side:
$$v+x\frac{dv}{dx}=v{\log }_{e}v+v$$
Subtracting $v$ from both sides:
$$x\frac{dv}{dx}=v{\log }_{e}v$$
Now, separate the variables:
$$\frac{dv}{v{\log }_{e}v}=\frac{dx}{x}$$
Integrating both sides:
$$\int \frac{dv}{v{\log }_{e}v}=\int \frac{dx}{x}$$
For the left-hand integral, substitute $u={\log }_{e}v$, which gives $du=\frac{1}{v}dv$:
$$\int \frac{du}{u}={\log }_{e}x+C_1$$
$${\log }_{e}u={\log }_{e}x+{\log }_{e}C$$
$${\log }_e\left({\log }_{e}v\right)={\log }_e\left(Cx\right)$$
Taking the exponential of both sides yields:
$${\log }_{e}v=Cx$$
Substituting back $v=\frac{y}{x}$:
$${\log }_e\left(\frac{y}{x}\right)=Cx$$
Thus, the general solution is indeed ${\log }_e\left(\frac{y}{x}\right)=Cx$. This makes Statement (C) true.

Step 3: Find the Particular Solution (Option E)
We are given the initial condition $y\left(1\right)=1$, meaning $y=1$ when $x=1$.
Substitute these values into the general solution:
$${\log }_e\left(\frac{1}{1}\right)=C\left(1\right)$$
$${\log }_e\left(1\right)=C\Rightarrow C=0$$
Substitute $C=0$ back into the general solution:
$${\log }_e\left(\frac{y}{x}\right)=0$$
$$\frac{y}{x}=e^0=1\Rightarrow y=x$$
This matches the particular solution given in statement (E). Thus, Statement (E) is true.

Since statements (B), (C), and (E) are all correct, the correct choice is (B), (C) and (E) only.