## Thomas' Calculus 13th Edition

$y=\dfrac{xe^{x/2}}{2}+ce^{x/2}$
Re-write the given differential equation as: $y'-\dfrac{y}{2}=\dfrac{e^{x/2}}{2}$ The integrating factor is: $v(x)=e^{\int (-1/2) dx}=e^{-(x/2)}$ Now, we have $\int [(e^{-x/2})y)' dx=\int \dfrac{1}{2} dx$ or, $y=e^{x/2}(\dfrac{x}{2})+c$ Hence, the General solution is: $y=\dfrac{xe^{x/2}}{2}+ce^{x/2}$