For the differential equation $y\, dx-(x + 3y^2)dy = 0$, which of the following statements are true?

(A) It is a linear differential equation
(B) It is a homogenous differential equation
(C) Its general solution is $x = 3y^2+ Cy$: C is an arbitrary constant
(D) If $y(0) = 1$, then its particular solution is $x = 3y^2 - 1$

Choose the correct answer from the options given below:

(A) and (C) only

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

Given differential equation: $y \, dx - (x + 3y^2) \, dy = 0$

Rewriting in standard form:

$\frac{dx}{dy} = \frac{x + 3y^2}{y}$

This is a linear differential equation in $x$ with respect to $y$:

$\frac{dx}{dy} - \frac{1}{y}x = 3y$

This is a linear equation (first order, linear in $x$).

So, (A) is correct.

Now check if it's homogeneous:

Original form: $\frac{dx}{dy} = \frac{x + 3y^2}{y} = \frac{x}{y} + 3y$

This is not homogeneous because the RHS is not a function of $\frac{x}{y}$ alone.

So, (B) is false.

Now solve the equation:

$\frac{dx}{dy} - \frac{1}{y}x = 3y$

Integrating factor: $\mu(y) = e^{\int -\frac{1}{y} \, dy} = e^{-\ln y} = \frac{1}{y}$

Multiplying through by $\frac{1}{y}$:

$\frac{1}{y} \cdot \frac{dx}{dy} - \frac{1}{y^2}x = 3$

$\frac{d}{dy} \left(\frac{x}{y} \right) = 3$

Integrating both sides:

$\frac{x}{y} = 3y + C$

$x = 3y^2 + Cy$

So, (C) is correct.

For particular solution: $x = 3y^2 + Cy$, plug in $y(0) = 1$

$0 = 3(1)^2 + C(1) \Rightarrow C = -3$

So, particular solution: $x = 3y^2 - 3y$

But option (D) says $x = 3y^2 - 1$, so it's incorrect.

So, (D) is false.