The integrating factor of frac{dy}{dx} + y = | vedclass

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(C) Given the differential equation:$\frac{dy}{dx} + y = \frac{1+y}{x}$Rearranging the terms to the standard form $\frac{dy}{dx} + Py = Q$:$\frac{dy}{

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$\frac{e^{x}}{x}$

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(C) Given the differential equation:$\frac{dy}{dx} + y = \frac{1+y}{x}$Rearranging the terms to the standard form $\frac{dy}{dx} + Py = Q$:$\frac{dy}{dx} + y - \frac{y}{x} = \frac{1}{x}$$\frac{dy}{dx} + y(1 - \frac{1}{x}) = \frac{1}{x}$Here,$P = 1 - \frac{1}{x}$.The integrating factor ($I$.$F$.) is given by:$I.F. = e^{\int P dx} = e^{\int (1 - \frac{1}{x}) dx}$$I.F. = e^{x - \ln|x|} = e^{x} \cdot e^{-\ln|x|} = e^{x} \cdot \frac{1}{x} = \frac{e^{x}}{x}$

The integrating factor of $\frac{dy}{dx} + y = \frac{1+y}{x}$ is

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