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(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = 3x$.First,

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$y = x^2 + \frac{C}{x}$

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(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = 3x$.First,we find the integrating factor $(IF)$:$IF = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$.Multiplying the differential equation by the integrating factor $x$,we get:$x \frac{dy}{dx} + y = 3x^2$.This can be written as:$\frac{d}{dx}(xy) = 3x^2$.Integrating both sides with respect to $x$:$\int \frac{d}{dx}(xy) dx = \int 3x^2 dx$.$xy = x^3 + C$.Dividing by $x$,we get the general solution:$y = x^2 + \frac{C}{x}$.

The general solution of the differential equation $\frac{dy}{dx} + \frac{y}{x} = 3x$ is

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