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.

  • $\begingroup$ The asymptotic behaviour is just sinusoidal—in particular, the $e^{-t/2}$ term vanishes as $t\to \infty$, while the remaining term does not. $\endgroup$ – 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!

  • 2
    $\begingroup$ Did you miss the minus sign? $\endgroup$ – Ian May 18 '18 at 2:36
  • $\begingroup$ Whoops, I solved the equation myself and ended up with a different equation, but I somehow typed in OP's equation when writing my answer. Thanks for the catch! $\endgroup$ – ThatEvilChickenNextDoor May 18 '18 at 2:51
  • $\begingroup$ I'm guessing OP mistyped the equation. You'd need a decaying exponential for the question to make sense $\endgroup$ – Dylan May 18 '18 at 5:59
  • $\begingroup$ Sorry for the mistake. I understand this now. Thank you! $\endgroup$ – Safder May 18 '18 at 12:04
  • $\begingroup$ Why do you say that for $C=0$ the behavior is undefined ? We know its exact expression, which is that of a sinusoid. $\endgroup$ – Yves Daoust May 18 '18 at 12:10

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