The general solution of the differential equa | vedclass

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(B) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{3x^2}{1+x^3}$ and $Q = \frac{1}{1+x^3}$.First,we find the

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$y(1+x^3) = x + c$,where $c$ is a constant of integration.

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(B) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{3x^2}{1+x^3}$ and $Q = \frac{1}{1+x^3}$.First,we find the Integrating Factor ($I$.$F$.):$	ext{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 $y \cdot (	ext{I.F.}) = \int Q \cdot (	ext{I.F.}) dx + c$.Substituting the values:$y(1+x^3) = \int \left(\frac{1}{1+x^3}ight) \cdot (1+x^3) dx + c$$y(1+x^3) = \int 1 dx + c$$y(1+x^3) = x + c$.

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

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