Solve IVP using Laplace Transforms: $y'' + 4y' + 5y = \delta(t-1), y(0) = 0, y'(0) = 3$ I have the following problem
$$y'' + 4y' + 5y = \delta(t-1),\quad y(0) = 0,\quad y'(0) = 3$$
When transforming I get
$$s^2Y - sy(0) - y'(0) + 4(sY - y(0)) + 5Y = e^{-s} \\
Y(s^2 + 4s +5) = e^{-s} + 3 \\
Y(s) = \frac{e^{-s} + 3}{s^2 + 4s + 5}$$
Now, I rewrote $s^2 + 4s +5 = (s + 2)^2 + 1$ 
$$Y(s) = e^{-s}\left( \frac{1}{(s+2)^2 + 1} \right) + 
3\left( \frac{1}{(s+2)^2 + 1}\right)$$
And transforming back I use the fact that 
$e^{at}\sin{bt} = \mathcal{L}^{-1}\left(\frac{b}{(s-a)^2 + b^2}\right)$ thus I get
$$y(t) = \delta(t-1) e^{-2t}\sin{t} + 3e^{-2t}\sin{t}$$
But this I get something else when wolfram solves the equation. And when plotting my solution it seems wierd. Can anyone check if there's anything wrong here?
Sorry about the formatting, couldn't get it to line up..
 A: Let's start off from the last point where you were on the right track:
$$Y(s) = \color{red}{e^{-s}\left( \frac{1}{(s+2)^2 + 1} \right)} + 
\color{green}{3\left( \frac{1}{(s+2)^2 + 1}\right)}$$
The only mistake you made is when evaluating the Inverse Laplace transform of the red part, as shown below:


In this answer, we prove the following result (Result 27 on the Table of Laplace Transforms):
  $$\mathcal{L}\{u_c(t)\cdot f(t-c)\}=e^{-cs}F(s) \tag{1}$$
  Where $F(s)=\mathcal{L}\{f(t)\}$ and $u_c(t)=u(t-c)$. This is often referred to as the Second Shifting Theorem.

By definition, we must consider the following improper integral:
$$\begin{align} \mathcal{L}\{u_c(t)f(t-c)\}&=\int_0^{\infty} e^{-st}u_c(t)f(t-c)~dt\\&=\int_c^{\infty} e^{-st}f(t-c)~dt \end{align}$$
Using the change of variable $y=t-c$, we obtain:
$$\int_c^{\infty} e^{-st}f(t-c)~dt=\int_0^{\infty} e^{-s(y+c)}f(y)~dy=e^{-cs} \int_0^{\infty} e^{-sy}f(y)~dy=e^{-cs}F(s) \tag*{$\blacksquare$}$$

Using $(1)$, and selecting $c=1$, we clearly see that the Inverse Laplace transform of the red part is:
$$\mathcal{L}^{-1}\left\{\frac{e^{-s}}{(s+2)^2+1}\right\}=u_1(t)\cdot e^{-2(t-1)}\sin(t-1)=e^{2-2t} u(t-1)\sin(t-1)$$
The answer you obtain will indeed be the same as Wolfram Alpha's answer, namely:
$$y(t)=e^{2-2t}u(t-1)\sin(t-1)+3e^{-2t}\sin(t)$$
