system of 1st order ODE's

system of 1st order ODE's

Given is the system of first order ODEs with initial value:
$x'(t)=\left(\begin{matrix}2 & 0 & 0\\ 1 & \alpha & \beta\\ 3 & \gamma > &
>
\alpha\end{matrix}\right)x(t)+\left(\begin{matrix}e^{3t}\\e^{3t}\\-e^{3t}\end{matrix}\right),x(0)=\left(\begin{matrix}1\\2\\-2\end{matrix}\right)$
Determine $\alpha,\beta, \gamma\in\mathbb{Z}$ so that
$\left(\begin{matrix}0\\e^{3t}cos(t)\\-e^{3t}sin(t)\end{matrix}\right)$ is
a solution of the homogeneus system.
Determine the solution of the initial value problem, with the
$\alpha,\beta, \gamma\in\mathbb{Z}$ from question 1.
Find a solution to the initial value problem using the solution in
question 2.
$x'(t)=\left(\begin{matrix}2 & 0 & 0 & 0\\ 1 & \alpha & 0 &\beta\\ >
\frac{9}{(t+1)^{2}} & 0 & -\frac{3}{(t+1)} & 0\\ 3 & \gamma & 0 & >
\alpha\end{matrix}\right)x(t)+\left(\begin{matrix}e^{3t}\\e^{3t}\\0\\-e^{3t}\end{matrix}\right),x(0)=\left(\begin{matrix}1\\2\\3\\-2\end{matrix}\right)$
Notice that the matrix has non-constant coëfficiënts on the 3th row, and
column 3 of the matrix contains zeros except on the 3th place in this row.
My solution:
question 1:
$x'(t)=\left(\begin{matrix}2 & 0 & 0\\ 1 & \alpha & \beta\\ 3 & \gamma &
\alpha\end{matrix}\right)x(t)$
$x(t)=\left(\begin{matrix}0\\e^{3t}cos(t)\\-e^{3t}sin(t)\end{matrix}\right)\longrightarrow
x'(t)=\left(\begin{matrix}0\\3e^{3t}cos(t)-e^{3t}sin(t)\\-3e^{3t}sin(t)-e^{3t}cos(t)\end{matrix}\right)$
so
$x'(t)=A\cdot x(t)$
and
$0=2\cdot 0$
$3e^{3t}cos(t)-e^{3t}sin(t)=1\cdot 0+\alpha\cdot e^{3t}cos(t) -\beta\cdot
e^{3t}sin(t)$
$-3e^{3t}sin(t)-e^{3t}cos(t)=3\cdot 0 + \gamma\cdot e^{3t}cos(t) -
\alpha\cdot e^{3t}sin(t)$
That gives us: $\gamma =-1$, $\alpha=3$ and $\beta=1$
and
$A=\left(\begin{matrix}2 & 0 & 0\\ 1 & 3 & 1\\ 3 & -1 & 3\end{matrix}\right)$
Question 2:
First i find the solution to the homogeneous problem:
the eigenvalues and eigenvectors for the matrix A are:
$r_{1}=2$, $\varepsilon_{1} =\left(\begin{matrix}2\\
2\\-1\end{matrix}\right)$, $r_{2}=3-i$,
$\varepsilon_{2}=\left(\begin{matrix}0\\ 1\\i\end{matrix}\right)$,
$r_{3}=3+i$, $\varepsilon_{3}=\left(\begin{matrix}0\\
1\\-i\end{matrix}\right)$
Then i transform $\varepsilon_{2}e^{r_{2}t}$ and
$\varepsilon_{3}e^{r_{3}t}$ to real functions. The solution of the
homogeneous problem is:
$\left(\begin{matrix}x_{1}(t)\\x_{2}(t)\\x_{3}(t)\end{matrix}\right)=C1\cdot\left(\begin{matrix}2\\2\\-1\end{matrix}\right)\cdot
e^{2t}+C2\cdot\left(\begin{matrix}0\\cos(t)\\-sin(t)\end{matrix}\right)\cdot
e^{3t}+C3\cdot
\left(\begin{matrix}0\\sin(t)\\cos(t)\end{matrix}\right)\cdot e^{3t}$
How do i find the particular solution to this problem??
question 3:
How do i solve this problem?? It is obvious that i have to do something
with that 3th row, but still i do not have a clue...