IVP solved with Laplace transformation - mistake? I want to solve following IVP with Laplace transformation.
\begin{align}
x''(t) + 2x'(t) + x(t) &=
\begin{cases}
0, & t < 0\\
1, & t \in (0;2)\\
3, & t > 2
\end{cases}\\
x(0) &= 0\\
x'(0)&=0\\
\end{align}
I've transformed left side of the equation like this:
\begin{align}
\mathcal{L}(x''(t) + 2x'(t) + x(t)) = s^2X(s) - sx(0) - x'(0) + 2(sX(s) - x(0)) + X(s)
\end{align}
With conditions applied:
$$
s^2X(s)+2sX(s)+X(s)
$$
Then I transformed right side of the equation like this:
\begin{align}
\mathcal{L}(f(t)) &= F(s)\\
&= \int^0_{-\infty}0\cdot e^{-st} dt + \int^2_{0}1\cdot e^{-st} dt + \int^\infty_{2}3\cdot e^{-st} dt\\
&= 0 + \left[ - \frac{1}{s}e^{-st} \right]_{t=0}^2 + 3 \left[ - \frac{1}{s}e^{-st} \right]_{t=2}^\infty\\
&= \frac{1-e^{-2s}}{s} + 3 \frac{e^{-2s}}{s}\\
&= \frac{1+2e^{-2s}}{s}
\end{align}
So, now the transformed equation with applied conditions looks like this:
\begin{align}
s^2X(s)+2sX(s)+X(s) &= \frac{1+2e^{-2s}}{s}\\
X(s)&= \frac{1+2e^{-2s}}{s(s^2 +2s +1)}
\end{align}
Now I attempted to decompose the right side to partial fractions:
\begin{align}
X(s)&= \frac{1+2e^{-2s}}{s(s^2 +2s +1)}\\
&= (1+2e^{-2s})\frac{1}{s(s+1)^2}\\
&= (1+2e^{-2s})\left( \frac{A}{s} + \frac{B}{s+1} + \frac{C}{(s+1)^2} \right)\\
&= (1+2e^{-2s})\left( \frac{1}{s} - \frac{1}{s+1} - \frac{1}{(s+1)^2} \right)\\
&= \frac{1}{s} - \frac{1}{s+1} - \frac{1}{(s+1)^2} + 2\frac{1}{s}e^{-2s} - 2\frac{1}{s+1}e^{-2s} - 2\frac{1}{(s+1)^2}e^{-2s}
\end{align}
Now I used Inversed Laplace Transformation (u is Heaviside unit-step function):
$$
x(t) = 1- e^{-t} - te^{-t} + 2u(t-2) -2 e^{2-t}u(t-2) - 2e^{2-t}(t-2)u(t-2)
$$

The problem is, that this solution doesn't work for $t < 0$:
$x''(-1) = 2e$
$x'(-1) = -e$
$x(-1) = 1$
$x''(-1) + 2x'(-1) + x(-1) = 2e -2e + 1 = 1$, but $-1 < 0$, so the result should be 0.
Can you see, what I'm doing wrong? I've tried to check my steps with WolframAlpha, but still I don't see the error.
The other situations ($t \in (0;2)$ and $t > 2$) seem to work ok.
 A: $$
\frac{1}{s} - \frac{1}{s+1} - \frac{1}{(s+1)^2} + 2\frac{1}{s}e^{-2s} - 2\frac{1}{s+1}e^{-2s} - 2\frac{1}{(s+1)^2}e^{-2s}
$$
The inverse Laplace is :
$$
x(t) = u(t)- e^{-t}u(t) - te^{-t}u(t) + 2u(t-2) -2 e^{2-t}u(t-2) - 2e^{2-t}(t-2)u(t-2)
$$
Then
$$x(t) = 0 \forall t < 0 $$

$$
\int_{-\infty (or 0)}^{\infty} u(t) e^{-st} dt = \int_{0}^{\infty} u(t) e^{-st}dt =\int_{0}^{\infty} e^{-st} dt = \frac{e^{-st}}{-s}|_{0}^{\infty} = \frac{1}{s}
$$
$$
\rightarrow \mathcal{L^{-1}}\left\{\frac{1}{s}\right\} = u(t)
$$
you can do the same with the others.
Also you can check a Laplace transform table online to verify this.
A: I'm not entirely sure about the particular scenario you're working on, but the Laplace transform is widely used on functions $f(t)$ defined for $t\ge 0$ as
$$f(s) = \mathcal{L}[f(t)] = \int_0^\infty f(t)e^{-st}\ dt$$
Let me emphasize the $t\ge0$ part. It's not defined for $t<0$. That's the definition that most authors and computer solvers implement. 
Wolfram Mathworld discusses a bilateral laplace transform $\mathcal{L^{(2)}}[f(t)] = \int_{-\infty}^\infty f(t)e^{-st}\ dt$. I'm not too keen on the differences, but I'm betting the transforms look different.
It appears like you are using the traditional laplace transform, which works for $t\ge0$ but doesn't apply otherwise. What I would suggest is to solve the $t\ge0$ equation and $t<0$ cases using different methods. After all the $t<0$ case is a simple 2nd order homogeneous ODE:
$$x'' + 2x' + x = 0$$
