#### Answer

$y=e^{x/2}[\dfrac{x^2}{4}+c]$

#### Work Step by Step

Here, we have $y'-\dfrac{1}{2}y=\dfrac{x}{2}e^{(x/2)}$
The integrating factor is: $I=e^{\int (-1/2)dx}=e^{-x/2}$
Now, $e^{-x/2}[y'-\dfrac{y}{2}]=(e^{-x/2})[(\dfrac{x}{2})(e^{(x/2)})]$
This implies that
$\int [e^{-(x/2)}y]' =\int [(\dfrac{x}{2})(e^{(x/2)})] (e^{-x/2}) dx$
Hence, $y=e^{(x/2)}[\dfrac{x^2}{4}+c]$