For the differential equation y dx-(x + 3y^2)dy = 0 , which of the fo…

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

A first-order equation is called linear if it is linear in its dependent variable, which may be x or y. This equation is not linear in y, but it is linear when x is taken as a function of y:

    \(y\,dx = (x + 3y^2)\,dy\)

    \(\displaystyle \frac{dx}{dy} = \frac{x + 3y^2}{y}\)

    \(\displaystyle \frac{dx}{dy} - \frac{1}{y}\,x = 3y\)

This is a linear equation in x, \(\frac{dx}{dy} + Px = Q\), with \(P = -\frac{1}{y}\) and \(Q = 3y\).

    \(\displaystyle \text{I.F.} = e^{\int -\frac{1}{y}\,dy} = e^{-\log_e y} = \frac{1}{y}\)

    \(\displaystyle \frac{x}{y} = \int 3y\cdot\frac{1}{y}\,dy = 3y + C\)

    \(x = 3y^2 + Cy\)

The condition \(y(0) = 1\) means y = 1 when x = 0:

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

    \(x = 3y^2 - 3y\)

For homogeneity, write the equation as

    \(\displaystyle \frac{dy}{dx} = \frac{y}{x + 3y^2}\)

The denominator mixes a term of degree 1 (x) with a term of degree 2 (\(3y^2\)), so the right side is not a function of \(\frac{y}{x}\) alone.

So the equation is linear (in x), it is not homogeneous, its general solution is \(x = 3y^2 + Cy\), and with \(y(0) = 1\) the particular solution is \(x = 3y^2 - 3y\).

Checking the statements

(A) It is a linear differential equation — Correct. It is \(\frac{dx}{dy} - \frac{x}{y} = 3y\), linear in x.

(B) It is a homogeneous differential equation — Incorrect. \(x + 3y^2\) mixes degrees 1 and 2.

(C) Its general solution is \(x = 3y^2+ Cy\): C is an arbitrary constant — Correct. With I.F. \(\frac{1}{y}\), \(\frac{x}{y} = 3y + C\), so \(x = 3y^2 + Cy\).

(D) If \(y(0) = 1\), then its particular solution is \(x = 3y^2 - 1\) — Incorrect. x = 0, y = 1 gives C = −3, so \(x = 3y^2 - 3y\).