For the differential equation $ydx - (x + 3y^2)dy = 0$, 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:

We're given:

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

Let's rearrange by dividing through by $dy$:

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

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

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

Notice we're treating $y$ as the independent variable and $x$ as the dependent variable. Writing $\frac{dx}{dy}$ instead of $\frac{dy}{dx}$ gives us a clean, solvable form.

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Checking Statement (A):

A linear differential equation in $x$ has the form:

$\frac{dx}{dy} + P(y) \cdot x = Q(y)$

Our equation $\frac{dx}{dy} - \frac{1}{y} \cdot x = 3y$ matches this form exactly, with $P(y) = -\frac{1}{y}$ and $Q(y) = 3y$.

✅ Statement (A) is true.

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Checking Statement (B):

From the original form:

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

For a homogeneous equation, the RHS must be expressible purely as a function of $\frac{x}{y}$. The leftover $3y$ term prevents this. Also, in $x + 3y^2$, the term $x$ has degree $1$ and $3y^2$ has degree $2$ — different degrees confirm it is not homogeneous.

❌ Statement (B) is false.

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Solving the equation to verify (C) and (D):

From $\frac{dx}{dy} - \frac{x}{y} = 3y$, we find the integrating factor:

$IF = e^{\int -\frac{1}{y}\,dy} = e^{-\ln y} = \frac{1}{y}$

Applying the formula $x \cdot IF = \int Q(y) \cdot IF \, dy$:

$\frac{x}{y} = \int 3y \cdot \frac{1}{y}\,dy$

$\frac{x}{y} = \int 3\,dy$

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

$x = 3y^2 + Cy$

This matches Statement (C) exactly.

✅ Statement (C) is true.

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Checking Statement (D):

$y(0) = 1$ means when $x = 0$, $y = 1$. Substituting into $x = 3y^2 + Cy$:

$0 = 3(1)^2 + C(1)$

$C = -3$

So the particular solution is $x = 3y^2 - 3y$.

Statement (D) claims $x = 3y^2 - 1$. Checking: at $x = 0, y = 1$, this gives $0 = 3 - 1 = 2$, which is a contradiction.

❌ Statement (D) is false.

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Statements $\text{(A)}$ and $\text{(C)}$ are true.