The Integrating Factor (I.F.) of the differen | vedclass

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

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$\frac{1}{1+x^3}$

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(D) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$, where $P = -\frac{3x^2}{1+x^3}$ and $Q = \frac{\sin^2(x)}{1+x}$.The Integrating Factor $(I.F.)$ is given by the formula $I.F. = e^{\int P dx}$.Substituting the value of $P$:$I.F. = e^{\int -\frac{3x^2}{1+x^3} dx}$.Let $u = 1+x^3$, then $du = 3x^2 dx$. Thus, $\int \frac{3x^2}{1+x^3} dx = \int \frac{1}{u} du = \log|1+x^3|$.Therefore, $I.F. = e^{-\log(1+x^3)} = e^{\log(1+x^3)^{-1}} = (1+x^3)^{-1} = \frac{1}{1+x^3}$.

The Integrating Factor $(I.F.)$ of the differential equation $\frac{dy}{dx} - \frac{3x^2y}{1+x^3} = \frac{\sin^2(x)}{1+x}$ is:

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