Given the following IVP:
$$\ddot y+4y=\cos t-\cos t \cdot \theta(t-2\pi), y(0)=0, \dot y(0)=1$$ Check that $y(t)$ is continuous at $t=2\pi$.
Find the jump in $\ddot y(t)$ at $t=2\pi$ i.e find $\lim _{t\to2\pi^+}\ddot y(t)-\lim _{t\to2\pi^-}\ddot y(t)$.
I know how to get the answer for the first one by using the solution. Is there another way to do it?
And I need some explanation for the "jump". I asked a quite similar question here to this a few days ago but I don't to use the solution I've been given for finding the jump in $y\ddot (t)$. Thank you very much!
$$\ddot y+4y=\cos t-\cos t \cdot \theta(t-2\pi)$$
Because $y(t)$ is continuous at $t=2\pi$, we have
$$\lim _{t\to2\pi^+}\ddot y(t)-\lim _{t\to2\pi^-}\ddot y(t)=\lim _{c\to 0}\left(\ddot y(2\pi+c)-\ddot y(2\pi-c)\right)$$ $$=-\cos (2\pi) \cdot \lim _{c\to 0}(\theta(c)-\theta(-c))=-1 \cdot (1-0)=-1$$