Answer:
We have, \[\frac{dy}{dx}+y=\frac{1+y}{x}\] \[\Rightarrow \] \[\frac{dy}{dx}+y=\frac{1}{x}+\frac{y}{x}\] \[\Rightarrow \] \[\frac{dy}{dx}+y-\frac{y}{x}=\frac{1}{x}\] \[\Rightarrow \]\[\frac{dy}{dx}+y\left( 1-\frac{1}{x} \right)=\frac{1}{x}\] \[\Rightarrow \] \[\frac{dy}{dx}+y\left( \frac{x-1}{x} \right)=\frac{1}{x}\] Which is a linear differential equation of the form \[\frac{dy}{dx}+Py=Q,\] where \[P=\frac{x-1}{x}\] and \[Q=\frac{1}{x}\] Now, \[IF={{e}^{\int{Pdx}}}={{e}^{\int{\frac{x-1}{x}dx}}}={{e}^{\int{\left[ 1\,-\,\frac{1}{x} \right]}\,dx}}\] \[={{e}^{x\,-\,\log \,x}}={{e}^{x}}\cdot {{e}^{-\log \,x}}\] \[={{e}^{x}}\cdot {{e}^{\log \,{{x}^{-1}}}}={{e}^{x}}\cdot {{x}^{-1}}={{e}^{x}}.\frac{1}{x}\]
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