Since the conjugate roots are repeated, the general solution takes the form:
$\displaystyle y=c_1e^{-\frac{1}{2}+\frac{\sqrt{3}}{2}i}+c_2e^{-\frac{1}{2}-\frac{\sqrt{3}}{2}i}+c_3xe^{-\frac{1}{2}+\frac{\sqrt{3}}{2}i}+c_4xe^{-\frac{1}{2}-\frac{\sqrt{3}}{2}i}$
However, we don't like to see complex numbers in the solution, so we use
Euler's Formula $\displaystyle \bigg(e^{i\theta}=\cos(\theta)+i\sin(\theta)\bigg)$ to clean things up:
After some calculations, we find that our general solution is:
$\displaystyle y=e^{-\frac{1}{2}x}\left[c_1\cos\left(\frac{\sqrt{3}}{2}\right)+c_2\sin\lef t(\frac{\sqrt{3}}{2}\right)\right]+xe^{-\frac{1}{2}x}\left[c_3\cos\left(\frac{\sqrt{3}}{2}\right)+c_4\sin\lef t(\frac{\sqrt{3}}{2}\right)\right]$
$\displaystyle \implies \color{red}\boxed{y=e^{-\frac{1}{2}x}\left[\left(c_1+c_3x\right)\cos\left(\frac{\sqrt{3}}{2}\ right)+\left(c_2+c_4x\right)\sin\left(\frac{\sqrt{ 3}}{2}\right)\right]}$
To see how we go from $\displaystyle y=c_1e^{(\alpha+\beta i)x}+c_2e^{(\alpha-\beta i) x}\implies y=e^{\alpha x}\left[c_1\cos(\beta x)+c_2\sin(\beta x)\right]$, read post #3 in my
Diff Equ Tutorial.