Asymptotic behavior of solution to differential equation

I've just started my first class in ODE's and I'm kind of stuck with the idea of asymptotic behavior.

The problem as is follows: Find the general solution and use it to determine the asymptotic behavior for different values of a. $y(0) = a$ as $t\rightarrow$ $+\infty$.

$$y' - \frac{1}{2}y = 2\cos(t)$$

I've solved for the general solution: $$y = \frac{4}{5}(2\sin(t)+\cos(t)) + Ce^{-t/2}$$

Where do I go from here? Thank you for any guidance.

• The asymptotic behaviour is just sinusoidal—in particular, the $e^{-t/2}$ term vanishes as $t\to \infty$, while the remaining term does not. – Guillermo Angeris May 18 '18 at 2:39

EDIT: The general solution actually appears to be $y=\frac{4}{5}(2\sin{t}-\cos{t})+Ce^{t/2}$, which I used but then copied OP's general solution when typing my answer. Sorry!
From my reading of the question, you want to determine the behavior of $y(t)$ as $t \rightarrow +\infty$. Your general solution consists of two parts; the first: $$\frac{4}{5}(2\sin{t}-\cos{t})$$ is periodic, and can be ignored in the long run once we consider the second: $$Ce^{t/2}$$ which grows without bound. So, we can say that $y(t) \rightarrow +\infty$ when $C>0$, $y(t) \rightarrow -\infty$ when $C<0$, and its behavior is undefined when $C=0$. Now you just have to figure out what initial conditions $y(0)=a$ correspond to what values of $C$, which should be pretty simple. I hope I read your question correctly, and good luck in your ODE class!
• Why do you say that for $C=0$ the behavior is undefined ? We know its exact expression, which is that of a sinusoid. – Yves Daoust May 18 '18 at 12:10