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(B) Given differential equation is: $\left(\frac{1}{x^2}+x\right) \frac{d y}{d x}+3 y=1$ Dividing by $\left(\frac{1+x^3}{x^2}\right)$, we get: $\frac{

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$(3 y-1) x^3+3 y=c$

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(B) Given differential equation is: $\left(\frac{1}{x^2}+x\right) \frac{d y}{d x}+3 y=1$ Dividing by $\left(\frac{1+x^3}{x^2}\right)$, we get: $\frac{d y}{d x}+\frac{3 x^2}{1+x^3} y=\frac{x^2}{1+x^3}$ This is a linear differential equation of the form $\frac{d y}{d x}+P y=Q$, where $P=\frac{3 x^2}{1+x^3}$ and $Q=\frac{x^2}{1+x^3}$. The Integrating Factor $(IF)$ is given by: $IF = e^{\int P d x} = e^{\int \frac{3 x^2}{1+x^3} d x} = e^{\log(1+x^3)} = 1+x^3$. The general solution is $y \cdot IF = \int (Q \cdot IF) d x + c$. Substituting the values: $y(1+x^3) = \int \left(\frac{x^2}{1+x^3}\right)(1+x^3) d x + c$. $y(1+x^3) = \int x^2 d x + c$. $y(1+x^3) = \frac{x^3}{3} + c$. Multiplying by $3$: $3y(1+x^3) = x^3 + 3c$. $3y + 3yx^3 - x^3 = 3c$. $(3y-1)x^3 + 3y = C$ (where $C = 3c$).

The general solution of the differential equation $\left(\frac{1}{x^2}+x\right) \frac{d y}{d x}+3 y=1$ is

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