I solved this eq. using the Laplace Transform:
$\ddot y+4\dot y+13 y=\delta(t-2\pi)-\delta(t-7\pi)$
The sol. is:
$y(t)=\frac{1}{3} e^{2 t} (-e^{14 \pi} \theta(t-7\pi) sin(3 t)+e^{4 \pi} \theta(t-2 \pi) sin(3 t)+2 sin(3 t)+3 cos(3 t))$
or $y(t)=\begin{cases} e^{-2t}\cos {3t}+\frac{2}{3}e^{-2t}\sin{3t} \quad \mbox{if} \quad t<2\pi\\ e^{-2t}\cos {3t}+(\frac{2}{3}+\frac{1}{3}e^{4\pi})e^{-2t}\sin{3t}\quad \mbox{if} \quad2\pi<t<7\pi \\ e^{-2t}\cos {3t}+(\frac{2}{3}+\frac{1}{3}e^{4\pi}+\frac{1}{3}e^{14\pi})e^{-2t}\sin{3t}\quad \mbox{if} \quad t>7\pi \end{cases}$
Now I am being asked what is the "jump" in $\dot y(t)$ at $t=2\pi$ and $t=7\pi$. Could any one explain me a way to answer it? I know how to find the difference in $\dot y$ at a certain point but not by what extent.
They're asking for
$$\lim_{t \to 2 \pi^+} y'(t) - \lim_{t \to 2 \pi^-} y'(t)$$
and the same at $7 \pi$. However, you don't need the solution to calculate these, all you need to assume is that $y(t)$ and $y'(t)$ are bounded (or even just integrable). Rewrite the equation
$$y''(t) = -4y'(t) - 13y(t) + \delta(t-2\pi)-\delta(t-7\pi)$$
and integrate both sides around $2 \pi$:
$$\int_{2 \pi - c}^{2 \pi + c} y''(t) dt = \int_{2 \pi - c}^{2 \pi + c} -4y'(t) - 13y(t) + \delta(t-2\pi)-\delta(t-7\pi) dt$$
By the FTC, the left hand side is $y'(2 \pi+c)-y'(2 \pi-c)$, so the jump at $2 \pi$ will be the limit as $c \to 0$ of the right hand side. In other words
$$\lim_{c \to 0} y'(2 \pi + c)-y'(2 \pi - c) = \lim_{c \to 0} \int_{2 \pi - c}^{2 \pi + c} -4y'(t) - 13y(t) + \delta(t-2 \pi) - \delta(t - 7 \pi) dt$$
Using that $y'$ and $y$ are integrable along with properties of the delta function we get
$$\lim_{c \to 0} y'(2 \pi + c)-y'(2 \pi - c) = 0 + 0 + 1 + 0 = 1.$$
By looking at the term that survived, you can see the pattern: if the equation is linear, its order is $n$, and the right side contains $a \delta(t-b)$, then the $(n-1)$th derivative of the solution will have a jump of size $a$ at $b$.