Given
$$y'' + 4y' + 5y = H(t-3)e^{-2t}, t>0, y(0) = 1, y'(0)=2 $$
To solve this diff. equation using Laplace transform.
Seems very straightforward. On one side, we have
$$\mathscr{L}\{y''+4y'+5y\} = \mathscr{L}\{y''\} + 4\mathscr{L}\{y'\} +5\mathscr{L}\{y\}=\\
\left [s^2\mathscr{L}\{y\}-sy(0)-y'(0)\right ] + 4\left [s\mathscr{L}\{y\}-y(0)\right ] + 5\mathscr{L}\{y\} =\\
\mathscr{L}\{y\}\left (s^2+4s+5\right ) -sy(0) -4y(0) -y'(0) = \mathscr{L}\{y\}\left (s^2+4s+5\right ) -s-6 $$
and on the other
$$\mathscr{L}\{H(t-3)e^{-2t}\} = \int_0^\infty e^{-(s+2)t}H(t-3)\mbox{d}t = \int_0^3 0\mbox{d}t + \int_3^\infty e^{-(s+2)t}\mbox{d}t = -\frac{1}{s+2}e^{-(s+2)t}\bigg\vert_3^\infty = \frac{e^{-3(s+2)}}{s+2} $$
Which ultimately yields:
$$\mathscr{L}\{y\} = \frac{s+6}{s^2+4s+5} +\frac{e^{-3(s+2)}}{(s+2)(s^2+4s+5)} $$
Since $\mathscr{L}$ is a linear operator, I take its inverse is also linear. Finding the inverse of the first summand is a piece of cake, however, how do we do the second one?
$$\mathscr{L}^{-1}\left\lbrace\frac{e^{-3(s+2)}}{(s+2)(s^2+4s+5)} \right\rbrace $$
Oh, missed a useful theorem: $$\forall r>0: \mathscr{L}\{f(t-r)\} = e^{-sr}\mathscr{L}\{f(t)\} $$
\begin{align} Y(s)&=\frac{s+6}{s^2+4s+5}+\frac{\mathrm e^{-3(s+2)}}{(s+2)(s^2+4s+5)}\\ &=\frac{s+2}{(s+2)^2+1}+\frac{4}{(s+2)^2+1}+\left[\frac{1}{s+2}-\frac{s+2}{(s+2)^2+1}\right]\mathrm e^{-3(s+2)}\\ \end{align} so you have $$ y(t)=\left[\cos t+4\sin t\right]\mathrm e^{-2t}+\left[H(t-3)-\cos (t-3)H(t-3)\right]\mathrm e^{-2t}$$