#### Answer

$y=\dfrac{xe^{x/2}}{2}+ce^{x/2}$

#### Work Step by Step

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:
$y=\dfrac{xe^{x/2}}{2}+ce^{x/2}$