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.
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!