We solve the initial value problem using the complex exponential version of the general solution.
Let $\omega=\frac{-1+\sqrt{3}i}{2}$ be one of the non-real roots of $x^3-1=0$. Then the other is $\omega^2$, and the general solution is $pe^x+qe^{\omega x}+re^{\omega^2 x}$. From the initial conditions we have $p+q+r=1$, $p+\omega q+\omega^2 r=0$ and $p+\omega^2q+\omega r=0$.
From the last two equations, by subtraction we get $q=r$. The first two equations now say that $p+2a=1$ and $p+(\omega+\omega^2)q=0$. But $\omega+\omega^2=-1$, so we get $p=q$ and therefore $p=q=r=\frac{1}{3}$. (Actually, the answer can be written down without any calculation if we are accustomed to cube roots of unity.)
If we want to, we can now express the $\frac{1}{3}(e^{\omega x}+e^{\omega^2 x})$ part in terms of $e^{-x/2}$ and sines and cosines. But we end up with something less attractive than $\frac{1}{3}\left(e^{x}+e^{\omega x}+e^{\omega^2 x}\right)$!