Answer
$y=\dfrac{xe^{x/2}}{2}+ce^{x/2}$
Work Step by Step
The standard form of the given equation is:
$y'-\dfrac{y}{2}=\dfrac{e^{x/2}}{2}$ ....(1)
The integrating factor is: $v(x)=e^{\int (-1/2) dx}=e^{-x/2}$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [(e^{-x/2})y)' dx=\int \dfrac{1}{2} dx$
or, $y=e^{x/2}(\dfrac{x}{2})+c$
Thus, the general solution is as follows:
$y=\dfrac{xe^{x/2}}{2}+ce^{x/2}$
