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(D) The given differential equation is $\frac{dy}{dx} + \frac{3x^2}{1 + x^3}y = \frac{\sin^2 x}{1 + x^3}$.This is a linear differential equation of th

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$y(1 + x^3) = \frac{x}{2} - \frac{1}{4}\sin 2x + c$

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(D) The given differential equation is $\frac{dy}{dx} + \frac{3x^2}{1 + x^3}y = \frac{\sin^2 x}{1 + x^3}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{3x^2}{1 + x^3}$ and $Q = \frac{\sin^2 x}{1 + x^3}$.First,we find the Integrating Factor $(I.F.)$:$I.F. = e^{\int P dx} = e^{\int \frac{3x^2}{1 + x^3} dx} = e^{\ln(1 + x^3)} = 1 + x^3$.The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.Substituting the values:$y(1 + x^3) = \int \frac{\sin^2 x}{1 + x^3} \cdot (1 + x^3) dx + c$$y(1 + x^3) = \int \sin^2 x dx + c$Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$:$y(1 + x^3) = \int \frac{1 - \cos 2x}{2} dx + c$$y(1 + x^3) = \frac{1}{2} \int (1 - \cos 2x) dx + c$$y(1 + x^3) = \frac{1}{2} (x - \frac{\sin 2x}{2}) + c$$y(1 + x^3) = \frac{x}{2} - \frac{\sin 2x}{4} + c$.

The solution of the differential equation $\frac{dy}{dx} + \frac{3x^2}{1 + x^3}y = \frac{\sin^2 x}{1 + x^3}$ is

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