Answer
$y=e^{x/2}[\dfrac{x^2}{4}+c]$
Work Step by Step
The standard form of the given equation is:
$y'-\dfrac{1}{2}y=\dfrac{x}{2}e^{x/2}$ ....(1)
The integrating factor is: $e^{\int (-1/2)dx}=e^{-x/2}$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [e^{-x/2}y]' =\int (\dfrac{x}{2}e^{x/2}) e^{-x/2} dx$ ...(2)
Thus, the equation (2) becomes:
$y=e^{x/2}[\dfrac{x^2}{4}+c]$
