Solving the differential equation $x'' + 5x' + 4x = e^{-t}$ I am asked to solve the following differential equation:

$x'' + 5x' + 4x = e^{-t}$

The procedures I follow are:
1- Find the general solution. For that,
\begin{align*}
r^2+5r+4 &= 0\\
(r+1)(r+4) &= 0\\
\\
y_g &= c_1 e^{-t} + c_2 e^{-4t}
\end{align*}
2- Find the particular solution. 
When doing that I called the particular solution $y_p = Ae^{-t}$, which means that $y_p' = - A e^{-t}$ and $y_p''= Ae^{-t}$. When inserting those on the given differential equation, that produces
\begin{align*}
Ae^{-t} - 5 Ae^{-t} + 4 Ae^{-t} &= e^{-t}\\
0 &= e^{-t}
\end{align*}
Where am I off here? Should I have called my particular solution something else?
Thank you.
Wolfram Alpha's output: link
 A: To solve:
$$x''(t)+5x(t)+4x(t)=e^{-t}$$
Use Laplace transform:
$$\mathcal{L}_t\left[x''(t)\right]_{\text{s}}+5\cdot\mathcal{L}_t\left[x'(t)\right]_{\text{s}}+4\cdot\mathcal{L}_t\left[x(t)\right]_{\text{s}}=\mathcal{L}_t\left[e^{-t}\right]_{\text{s}}$$
Now use:


*

*$$\mathcal{L}_t\left[x''(t)\right]_{\text{s}}=\text{s}^2\text{X}(\text{s})-\text{s}x(0)-x'(0)$$

*$$\mathcal{L}_t\left[x'(t)\right]_{\text{s}}=\text{s}\text{X}(\text{s})-x(0)$$

*$$\mathcal{L}_t\left[x(t)\right]_{\text{s}}=\text{X}(\text{s})$$

*$$\mathcal{L}_t\left[e^{-t}\right]_{\text{s}}=\frac{1}{1+\text{s}}$$


So, we get:
$$\text{s}^2\text{X}(\text{s})-\text{s}x(0)-x'(0)+5\cdot\left(\text{s}\text{X}(\text{s})-x(0)\right)+4\text{X}(\text{s})=\frac{1}{1+\text{s}}$$
Solving $\text{X}(\text{s})$:
$$\text{X}(\text{s})=\frac{1+x(0)(\text{s}^2+6\text{s}+5)+x'(0)(1+\text{s})}{(1+\text{s})^2(\text{s}+4)}$$
With inverse Laplace transform, we find:
$$x(t)=\frac{e^{-4t}\left(1-3x(0)-3x'(0)+e^{3t}\left(3t+12x(0)+3x'(0)-1\right)\right)}{9}$$
A: If $P(D) = (D+1)(D+4)$ the particular solution will be
$$ y_{p} = \frac{te^{-t}}{P(-1)}$$
To see why recall the exponential shift rule for the differentiation operator:
$$ P(D)e^{at}u(t) = e^{at}P(D+a)u(t)$$
where $u(t)$ is some function of $t$
Hence,
$$ (D+1)(D+4)e^{-t}t = e^{-t}D(D+3)t = 3e^{-t} $$
We almost obtain $e^{-t}$ in the right hand side.
Now note that if $P(D) = (D+1)(D+4)$
$$ P'(D) = (D+1)+(D+4)$$
If we evaluate in $D=-1$
$$ P'(-1) = 3$$
So
$$ (D+1)(D+4)\frac{e^{-t}t}{P'(-1)} = \frac{e^{-t}}{3} D(D+3)t = e^{-t} $$
So our particular solution will be
$$ y_{p} = \frac{te^{-t}}{3}$$
In general for
$$ y'' + Ay' + By = e^{ax} \quad \textrm{ and } P(a) =0 $$
we have the particular solution
$$ y_{p} = \frac{te^{at}}{P'(a)} \quad P'(a) \neq 0 $$
