The solution of the differential equation fra | vedclass

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(D) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{3x^2}{1+x^3}$ and $Q = \fr

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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 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 \left( \frac{\sin^2 x}{1+x^3} \right) (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}x - \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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