How can you solve $y''=\delta(t-c)$? So if $c\geq0$ and $a,b\in\mathbb{Z}$ with $y(0)=a$, $y'(0)=b$ how can I solve $y''=\delta(t-c)$ with Laplace transforms?
What I have so far is 
$$s^2Y(s)-sy(0)-y'(0)=e^{-cs} \\
s^2Y(s)-sa-b=e^{-cs} \\
Y(s) = \frac{e^{-cs}}{s^2}+\frac{a}{s}+\frac{b}{s^2} \\
y(t) = \begin{cases} a+bt \mbox {   if 0}  \leq \mbox{ t < c} \\
\mbox{something} \mbox{ if c}\leq\mbox{ t }
\end{cases}
$$
I am having trouble finding the inverse Laplace transform of $\frac{e^{-cs}}{s^2}$. Any help would be greatly appreciated.
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$\ds{{\rm y}\pars{0} = a\,,\quad{\rm y}'\pars{0} = b\,,\quad c > 0}$.

$$
{\rm y}''\pars{t} = \delta\pars{t - c}\ \imp\
{\rm y}'\pars{t} - {\rm y}'\pars{0} = \int_{0}^{t}\delta\pars{\tau - c}\,\dd\tau 
\ \imp\ {\rm y}'\pars{t} = b + \Theta\pars{t - c}
$$
where $\ds{\Theta\pars{t}}$ is the
Heaviside Step Function.

$$
{\rm y}\pars{t} - {\rm y}\pars{0}
=\int_{0}^{t}\bracks{b + \Theta\pars{\tau - c}}\,\dd\tau
=bt + \Theta\pars{t - c}\int_{c}^{t}\dd\tau
=bt + \Theta\pars{t - c}\pars{t - c}
$$

$$
\bbox[15px,border:1px solid black]{\ds{%
{\rm y}\pars{t} =a + bt + \Theta\pars{t - c}\pars{t - c}}}
$$
A: You can actually do this one directly, using the definition of the dirac delta. Integrate both sides over $(0,t)$.
$$ \int_0^t y''(\tau) \,d\tau = \int_0^t \delta(\tau-c)\,d\tau. $$
This evaluates to
$$ y'(t)-y'(0) = \begin{cases} 0 & t<c \\ 1 & t \geq c. \end{cases} $$
Integrating again, we have
$$ y(t)-y(0)-ty'(0) = \begin{cases} 0 & t < c \\ t-c & t \geq c. \end{cases} $$
Bringing things around and using $a=y(0)$ and $b=y'(0)$, you obtain
$$ y(t) = a+bt+(t-c)u(t-c). $$
Here, we use $u(t)$ to mean the unit heaviside function, which is 0 for $t<0$ and 1 for $t>0$. That should also give you your hint if you need to use Laplace transforms instead; Laplace transforms take heaviside functions into exponential shifts -- check the Wikipedia page on Laplace transforms for examples.
[edit] replaced typo of hessian with heaviside.
