Integrating factor of the differential equati | vedclass

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(D) Given differential equation is $\frac{dy}{dx} + y = \frac{1+y}{x}$.Rearranging the terms,we get $\frac{dy}{dx} + y = \frac{1}{x} + \frac{y}{x}$.Gr

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

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(D) Given differential equation is $\frac{dy}{dx} + y = \frac{1+y}{x}$.Rearranging the terms,we get $\frac{dy}{dx} + y = \frac{1}{x} + \frac{y}{x}$.Grouping the $y$ terms,we have $\frac{dy}{dx} + y - \frac{y}{x} = \frac{1}{x}$,which simplifies to $\frac{dy}{dx} + \left(1 - \frac{1}{x}\right)y = \frac{1}{x}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1 - \frac{1}{x}$ and $Q = \frac{1}{x}$.The integrating factor ($I$.$F$.) is given by $e^{\int P dx}$.$I$.$F$. $= e^{\int (1 - \frac{1}{x}) dx} = e^{x - \log x} = e^x \cdot e^{-\log x} = e^x \cdot e^{\log(x^{-1})} = e^x \cdot \frac{1}{x} = \frac{e^x}{x}$.

Integrating factor of the differential equation $\frac{dy}{dx} + y = \frac{1+y}{x}$ is

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