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(B) Given the differential equation: $(x + 3y^3) \frac{dy}{dx} = y$.
Rearranging the equation,we get: $\frac{dx}{dy} = \frac{x + 3y^3}{y} = \frac{x}{

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

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(B) Given the differential equation: $(x + 3y^3) \frac{dy}{dx} = y$.
Rearranging the equation,we get: $\frac{dx}{dy} = \frac{x + 3y^3}{y} = \frac{x}{y} + 3y^2$.
This can be written as: $\frac{dx}{dy} - \frac{1}{y}x = 3y^2$.
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = -\frac{1}{y}$ and $Q = 3y^2$.
The Integrating Factor $(I.F.)$ is given by: $I.F. = e^{\int P dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = e^{\ln(y^{-1})} = \frac{1}{y}$.
The general solution is given by: $x(I.F.) = \int (Q 	imes I.F.) dy + C$.
Substituting the values: $x \cdot \frac{1}{y} = \int (3y^2 \cdot \frac{1}{y}) dy + C$.
$\frac{x}{y} = \int 3y dy + C$.
$\frac{x}{y} = \frac{3y^2}{2} + C$.
Multiplying by $y$: $x = \frac{3y^3}{2} + Cy$.

Find the general solution of the differential equation: $(x + 3y^3) \frac{dy}{dx} = y$ where $y > 0$.

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