Which of the following statements is/are true?

(A) $(\tan^{-1} y - x)dy = (1 + y^2)dx$ is a differential equation where variables are separable.

(B) $(1 + x^2)dy + 2xydy = \cot x \ dx (x \neq 0)$ is a first order linear differential equation.

(C) $(4x + 6y + 5)dy - (3y + 2x + 4)dx = 0$ is not a homogeneous differential equation.

(D) $(xy)dx - (x + y^2)dy = 0$ is a homogeneous differential equation.

Choose the correct answer from the options given below:

Checking statement (A): $(tan^{-1} y - x)dy = (1 + y^2)dx$

A differential equation has separable variables if it can be written as $f(y)dy = g(x)dx$ where all y-terms are on one side and all x-terms on the other.

$(tan^{-1} y - x)dy = (1 + y^2)dx$

The left side has $(tan^{-1} y - x)$ which contains both $x$ and $y$ mixed together. The variables cannot be separated into pure y-terms and pure x-terms.

Statement (A) is FALSE.

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Checking statement (B): $(1 + x^2)dy + 2xydy = \cot x dx$

Note: This likely contains a typo and should be $(1 + x^2)dy + 2xydx = \cot x dx$

$(1 + x^2)dy = \cot x dx - 2xydx$

$(1 + x^2)dy = (\cot x - 2xy)dx$

$\frac{dy}{dx} = \frac{\cot x - 2xy}{1 + x^2}$

$\frac{dy}{dx} = \frac{\cot x}{1 + x^2} - \frac{2xy}{1 + x^2}$

$\frac{dy}{dx} + \frac{2x}{1 + x^2}y = \frac{\cot x}{1 + x^2}$

This matches the first order linear form $\frac{dy}{dx} + P(x)y = Q(x)$ with:

$P(x) = \frac{2x}{1 + x^2}$ and $Q(x) = \frac{\cot x}{1 + x^2}$

Both are functions of $x$ only.

Statement (B) is TRUE.

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Checking statement (C): $(4x + 6y + 5)dy - (3y + 2x + 4)dx = 0$

For $Mdx + Ndy = 0$, the equation is homogeneous if both $M$ and $N$ are homogeneous functions of the same degree.

$M = -(3y + 2x + 4) = -2x - 3y - 4$

$N = 4x + 6y + 5$

Terms like $2x, 3y, 4x, 6y$ have degree 1.

Terms like $4$ and $5$ have degree 0 (constants).

Since there are mixed degrees (degree 1 and degree 0), these are not homogeneous functions.

The equation is NOT homogeneous.

Statement (C) is TRUE.

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Checking statement (D): $(xy)dx - (x + y^2)dy = 0$

$M = xy$ (degree 2: power of $x$ + power of $y$ = 1 + 1 = 2)

$N = -(x + y^2)$

In $N$, term $x$ has degree 1 and term $y^2$ has degree 2.

Since $N$ has mixed degrees (1 and 2), $N$ is not a homogeneous function.

For the equation to be homogeneous, both $M$ and $N$ must be homogeneous of the same degree.

Statement (D) is FALSE.

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(A) FALSE

(B) TRUE

(C) TRUE

(D) FALSE

Only statements (B) and (C) are true.