Laplace transform of the square wave to solve PDE Solve $$y'' + 3y' +2y = r(t)$$ given $y(0)=0$ and $y'(0) = 0$ where $r(t)$ is the square wave, $$r(t) = u(t-1) - u(t-2)$$

I'm just going to type out the answer as I read it and tell you which parts I understand and which parts I have trouble intuitively grasping, I hope this is OK.

SOLUTION
Using Laplace transforms, 
$$
\begin{array}
\ L[y'' +3y'+2y] = L[r(t)]\\
\implies L[y'' +3y' +2y] = L[u(t-1) - u(t-2)]\\
\implies [s^2 Y(s) - s y(0) - y'(0)] + 3[s Y(s) - y(0)] +2 Y(s) = \color{green}{{1 \over s} [e^{-s} - e^{-2s}]}\\
\end{array}
$$

The last part in $\color{green}{\text{green}}$ I guess comes from this result for the $\color{green}{\text{Dirac's Delta Function}}$:
  $$L[\delta (t-a)] = e^{-as}$$
  (Because the square wave is a step function?)

Now,
$$
\ Y(s) [s^2 +3s +2] = {1 \over s} [e^{-s} - e^{-2s}] = \frac{1}{(s{^2} +3s +2 s)s} [e^{-s} - e^{-2s}]\\
$$
$$F(s)= \frac{1}{(s{^2} +3s +2 s)s} = {1 \over 2s} - {1 \over {s+1}} + {1 \over s(s+2)} $$
$$ f(t) = L^{-1}[F(s)] = L^{-1}[{1 \over 2s} - {1 \over {s+1}} + {1 \over s(s+2)}] \implies f(t) = {1 \over 2} - e^t + {1 \over 2} e^{-2t}$$

Up to this point I think I can grasp what's going on. But after this, I'm not 100% sure I follow:

$$ Y(s) = L^{-1} [ F(s) e^{-s} - F(s) e^{-2s}] = \color{blue}{f(t-1)u(t-1) - f(t-2)u(t-2)}$$
$$ 
\color{red}{
Y(s) =
\begin{cases}
0, &0 < t< 1 \\[2ex]
{1 \over 2} - e^{-(t-1)} + {1 \over 2} e^{-2(t-1)}, &1< t<2 \\[2ex]
{1 \over 2} e^{-(t-1)} - {1 \over 2} e^{-2(t-1)} - {1 \over 2} e^{(t-2)} + {1 \over 2} e^{2(t-2)}, &t>2 \\[2ex]
\end{cases}}
$$

I have absolutely no idea how $\color{blue}{this}$ became  $\color{red}{this}$ 
I've looked at this result for a few days now but can't seem to get the intuition right. That's why I came here. I need to understand how this answer became this set of cases. I would appreciate your help very much!

 A: For $t<1$ we have $u(t-1)=0$ and for $t<2$ we have $u(t-2)=0$. So for $t<1$ we have $f(t-1)u(t-1)=0$ and $f(t-2)u(t-2)=0$ and then $y(t)=0$.
For $t> 1$ we have $u(t-1)=1$ and for $t<2$ we have $u(t-2)=0$. So for $1<t<2$ we have $f(t-1)u(t-1)=f(t-1)$ and $f(t-2)u(t-2)=0$ and then $$y(t)=f(t-1)={1 \over 2} - e^{-(t-1)} + {1 \over 2} e^{-2(t-1)}$$.
For $t> 2$ we have $u(t-1)=1$ and for $t>2$ we have $u(t-2)=1$. So for $t>2$ we have $f(t-1)u(t-1)=f(t-1)$ and $f(t-2)u(t-2)=f(t-2)$ and then $$y(t)=f(t-1)-f(t-2)={1 \over 2} - e^{-(t-1)} + {1 \over 2} e^{-2(t-1)}-\left[{1 \over 2} - e^{-(t-1)} + {1 \over 2} e^{-2(t-1)}\right]={1 \over 2} e^{-(t-1)} - {1 \over 2} e^{-2(t-1)} - {1 \over 2} e^{(t-2)} + {1 \over 2} e^{2(t-2)}$$.
A: $$ y(t) = L^{-1} [ F(s) e^{-s} - F(s) e^{-2s}] = (t-1)u(t-1) - f(t-2)u(t-2)$$
$u(t-t_0)$ is the unit step function. It has the value $0$ for all values of $t<t_0$ and has the value $1$ for all values of $t>t_0$. Using this definition you can see that for $0<t<1$ all unit step functions vanish. Resulting in $y(t)=0$.
For $1<t<2$ only $f(t-1)$ will appear. For $t>2$ $f(t-1)-f(t-2)$ is the result.
