Calculating the particular solution of the following nonhomogeneous system The name of the game is systems of differential equations and matrix exponentials. I have the following problem: Apply the formula $\vec x(t) = e^{At} \vec c + e^{At} \int e^{At} \vec f(t) dt$ (where $A$ is the coefficient matrix) with $\vec c = 0$ to find a particular solution of the following nonhomogeneous system,
$$
      \vec x ^\prime = \begin{bmatrix}
        1 & 1 & 0 \\
        0 & 1 & 1 \\
        0 & 0 & 1 \\
        \end{bmatrix} \vec x + 
        \begin{bmatrix}
        6\\
        12\\
        24\\
        \end{bmatrix}te^t
$$
I need to find $e^{At}$, which can be calculated by $e^{At} = P\Lambda_t P^{-1}$, where the column vectors of $P$ are the eigenvectors associated with each eigenvalue of $A$, and $\Lambda_t$ is the diagonal matrix with the diagonal entries of the form $e^{\lambda t}$ for each eigenvalue $\lambda$ of $A$. E.g., 
$$
        \Lambda_t = \begin{bmatrix}
        e^{\lambda_1 t} & 0 & 0 \\
        0 & e^{\lambda_2 t} & 0 \\
        0 & 0 & e^{\lambda_3 t} \\
        \end{bmatrix}
$$
Upon calculating $\det (A - I \lambda)$, we find (unfortunately) that the characteristic equation of $A$ is $(1 - \lambda)^3$. So our only eigenvalue is $\lambda = 1$ with a multiplicity of three and a defect of two. For the sake of time and my job in the morning, I will cut to the chase:
Question 1: The only eigenvalue of $A$ is $\lambda = 1$ of multiplicity three. Would that mean 
$$
        \Lambda _t = \begin{bmatrix}
        e^t & 0 & 0 \\
        0 & e^t & 0 \\
        0 & 0 & e^t \\
        \end{bmatrix}
$$
Question 2: To accommodate the multiplicity three eigenvalue, we must form a three-chain $\{ v_1, v_2, v_3 \}$ found by solving $(A - \lambda)^3 v_3 = 0, v_2 = (A - \lambda)v_3, v_1 = (A - \lambda)v_2$. I began with 
$$
        v_3 = \begin{bmatrix}
        1\\
        1\\
        1\\
        \end{bmatrix}
$$
and end with
$$
        P = \begin{bmatrix}
        1 & 1 & 1 \\
        0 & 1 & 1 \\
        0 & 0 & 1 \\
        \end{bmatrix}
$$
Is this value for $P$ correct? Note that the three vectors imply that our solution is of the form
$$x_1 (t) = v_1 e^t$$
$$x_2 (t) = (v_1 t + v_2) e^t$$
$$x_3 (t) = (\frac{1}{2} v_1 t^2 + v_2 t + v_3) e^t$$
I tried using $v_1, v_1 t + v_2,$ and $\frac{1}{2} v_1 t^2 + v_2 t + v_3$ as my eigenvectors (and thus column vectors for $P$), but I came out to the same answer as below. Using my values of $P$ and $\Lambda _t$ in the formula stated in the problem, I do not get the right answer and instead get
$$
        \begin{bmatrix}
        3t^2 e^t\\
        6t^2 e^t\\
        12t^2 e^t\\
        \end{bmatrix}
$$
whereas the correct answer is 
$$
        e^t \begin{bmatrix}
        3t^2 + 2t^3 + t^4\\
        6t^2 + 4t^3\\
        12t^2\\
        \end{bmatrix}
$$
Any help would be very much appreciated.
 A: Solve:
$$ x'(t) = \begin{pmatrix}
        1 & 1 & 0 \\
        0 & 1 & 1 \\
        0 & 0 & 1 \\
        \end{pmatrix} x(t) + 
        \begin{pmatrix}
        6\\
        12\\
        24\\
        \end{pmatrix}te^t$$
Method 1: Decoupled $z$
Notice that the third equation is decoupled from the other two.
Based on this, we can easily solve using any of various methods:
$$z' = z + 24 t e^t \implies z(t) = e^t( c_3 + 12 t^2)$$
Substituting this into the second equation, we have:
$$y' = y + z + 12 t e^t = y + e^t( c_3 + 12 t^2) + 12 t e^t \implies y(t) = e^t( c_2 + c_3 t + (4 t + 6) t^2)$$
Repeating this process for the first equation yields:
$$x(t) = e^t \left( c_1 + c_2 t + \dfrac{1}{2} c_3 t^2 + (t^2 + 2 t + 3)t^2\right)$$
Method 2: Homogeneous and particular - we have the matrix:
$$\begin{pmatrix}
        1 & 1 & 0 \\
        0 & 1 & 1 \\
        0 & 0 & 1 \\
        \end{pmatrix}$$
We have the eigenvalues $\lambda_{1,2,3} = 1$.
For the first eigenvalue, we solve $[A - I]v_1 = 0 \implies v_1 = (1, 0, 0)$
We cannot find two more linearly independent eigenvectors given this is a deficient matrix, so we find generalized ones.
For a second eigenvector, we will solve $[A - I]v_2 = v_1 \implies v_2 = (0, 1, 0)$.
For the third eigenvector, we will solve $[A - I]v_3 = v_2 \implies v_3 = (0, 0, 1)$.
These three eigenvectors make up your $P$.
For the homogeneous solution, we can now write:
$$x_h(t) = e^t \left(c_1 v_1 + c_2 (v_1 t + v_2) + c_3\left( \dfrac{1}{2} v_1  t^2 + v_2 t + v_3\right) \right)$$
Thus:
$$x_h(t) = e^t\left( c_1 \begin{pmatrix}  1\\ 0\\ 0\\ \end{pmatrix} + c_2 \left(\begin{pmatrix}  1\\ 0\\ 0\\ \end{pmatrix} t + \begin{pmatrix}  0\\ 1\\ 0\\ \end{pmatrix}\right) + c_3\left(\dfrac{1}{2} \begin{pmatrix}  1\\ 0\\ 0\\ \end{pmatrix}t^2 + \begin{pmatrix}  0\\ 1\\ 0\\ \end{pmatrix} t + \begin{pmatrix}  0\\ 0\\ 1\\ \end{pmatrix}\right)  \right)$$
This is the method you refer to in your link.
Can you continue to find the particular solution, $x_p(t)$ from the link I provided (Method I gives a complete answer for comparison)?
Update 
Method 3: Matrix Exponential (see Example 3)
To be clear, from the work above, we have $\displaystyle e^{A t} = e^t \begin{pmatrix}
 1 &  t & \dfrac{t^2}{2} \\
 0 & 1 &  t \\
 0 & 0 & 1 \\
\end{pmatrix}$. 
You want to follow example 3 in my link, where, for some initial condition $x(t_0)$ (which is not specified and can be any $t_0$), we would write the final solution as:
$$\displaystyle x(t) = e^{A(t-t_0)}x(t_0) +  \int_{t_0}^t e^{A(t-s)}~f(s)~ds,$$
where $f(s) = \begin{pmatrix}
        6\\
        12\\
        24\\
        \end{pmatrix}se^s$
It is also worth noting that there are many ways to solve such problems, these are only three examples (my link has two others), for example, see Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later
