Why is the complementary solution in trigonometric terms? So far the complementary solution (that is the solution to the associated homogeneous equation) has only been in the form: $$y(t)=c_1e^{at}+c_2e^{bt}$$ Where $a$ and $b$ are obtained from solving the characteristic equation. 

But for: $$4y^{\prime\prime}+y=e^{-2t}\sin({\frac{t}{2}}) + 6t\cos({\frac{t}{2}})$$ The complimentary solution is given as: $$y(t)=c_1\cos(\frac{t}{2}) +c_2\sin(\frac{t}{2})$$

Edit: The characteristic equation to the equation give above is: $$y^2+\frac{1}{4}=0$$ But where does the $\sin\theta$ and $\cos\theta$ come from?
How do you solve for $a$ and $b$? You solve $$4x^2+1=0$$ which has solutions $$x=\pm\frac i2$$ with $i=\sqrt{-1}$. So the complimentary solution is $$Ae^{it/2}+Be^{-it/2}$$
You can now use Euler's formula $$e^{i\theta}=\cos\theta+i\sin\theta$$ to write the complimentary solution as $$\begin{align}Ae^{it/2}+Be^{-it/2}&amp;=A\left(\cos\frac{\theta} 2+i\sin\frac{\theta} 2\right)+B\left(\cos\frac{\theta} 2-i\sin\frac{\theta} 2\right)\\&amp;=(A+B)\cos\frac{\theta} 2+i(A-B)\sin\frac{\theta} 2\end{align}$$
This is your equation if you just write $c_1=A+B$ and $c_2=i(A-B)$.
Well you see that in your equation when you make a characteristic equation of it you'll get $\lambda=0$
So it becomes $e^0=1$

Which is a constant