solve $y''-4y=\delta(t-1)$ with initial conditions $y(0)=0, \; y'(0)=1$ using Laplace transforms I took the Laplace transform and solved for $Y$ which resulted in $Y=\frac{1+e^{-s}}{s^2-4}$. I began to break up the problem separating the result into two equations but the fact that there is a $1$ added to the $e^{-s}$ is messing me up. The answer I got was $y(t)= \frac{1}{8} u_1(t)e^{-2t}$ but I'm not sure if that is right. 
 A: I think your answer is wrong. Note
\begin{eqnarray}
y(t)&=&L^{-1}\{\frac{1+e^{-s}}{s^2-4}\}\\
&=&L^{-1}\{\frac{1}{s^2-4}\}+L^{-1}\{\frac{e^{-s}}{s^2-4}\}\\
&=&\frac{1}{2}\sinh(2t)+\frac{1}{2}u_1(t)\sinh(2(t-1))
\end{eqnarray}
Here we used $L^{-1}\{e^{-as}F(s)\}=u_a(t)f(t-a)$
A: From the Laplace transform, we obtain
$$
s^2Y(s) - 1 - 4Y(s) = e^{-s}\iff Y(s) = \frac{1+e^{-s}}{s^2-4}
$$
as you have found. You can do this with tables and convolution theorems by writing the RHS as
$$
\frac{1}{s^2 - 4} + \frac{e^{-s}}{s^2 - 4}\tag{1}
$$
which I will leave this method to you. 
However, I will present an alternate method via the use of the inverse Laplace transform, $\mathcal{L}^{-1}\{F(s)\}(t)=\frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty}F(s)e^{st}ds$. In both terms of equation $(1)$, we have poles at $s=\pm 2i$ of order one so they are simple poles.
\begin{align}
y(t) & = \frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{st}}{s^2-4}ds+\frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{s(t-1)}}{s^2 - 4}ds\\
&=\lim_{s\to 2i}(s-2i)\frac{e^{st}}{s^2-4}+\lim_{s\to -2i}(s+2i)\frac{e^{st}}{s^2-4}+\frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{s(t-1)}}{s^2 - 4}ds\\
&= \frac{\sinh(2t)}{2}+\frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{s(t-1)}}{s^2 - 4}ds\\
&=\frac{\sinh(2t)}{2} + \lim_{s\to 2i}(s-2i)\frac{e^{s(t-1)}}{s^2 - 4}+\lim_{s\to -2i}(s+2i)\frac{e^{s(t-1)}}{s^2 - 4}\tag{2}
\end{align}
For equation $(2)$, we need to worry about convergence of the exponential term. The exponential can blow up if $s(t-1)>0$; therefore, we require that $s(t-1)<0\iff t < 1$. Therefore, in the time domain, we need the integral to be zero when $t<1$. We can achieve this with the shifted unit step, $\mathcal{U}(t-1)=\begin{cases}0,&t<1\\1,&t>1\end{cases}$. Let's adjust equation $(2)$ with this new information.
\begin{align}
y(t) &= \frac{\sinh(2t)}{2}+ \mathcal{U}(t-1)\lim_{s\to 2i}(s-2i)\frac{e^{s(t-1)}}{s^2 - 4}+\mathcal{U}(t-1)\lim_{s\to -2i}(s+2i)\frac{e^{s(t-1)}}{s^2 - 4}\\
&=\frac{1}{2}\bigl[\sinh(2t)-\sinh(2-2t)\mathcal{U}(t-1)\bigr]
\end{align}
