Chapter 9: First-Order Differential Equations - Practice Exercises - Page 557: 9

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

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