Solving an ODE system with duplicate eigenvalues I'm supposed to solve the following ODE system:  
$\left\{
 \begin{array}{ll}
  \frac{dx}{dt}=11x+6y+18z\\
  \frac{dy}{dt}=9x+8y+18z\\
        \frac{dz}{dt}=-9x-6y-16z
 \end{array}
\right.$  
Here is what I've done so far:
I begin by changing the problem into matrix form:  
$\left(\begin{matrix}
\frac{dx}{dt}\\
\frac{dy}{dt}\\
\frac{dz}{dt}
\end{matrix}\right)$=$\left(\begin{matrix}
11&6&18\\
9&8&18\\
-9&-6&-16
\end{matrix}\right)$
$\left(\begin{matrix}
x\\
y\\
z
\end{matrix}\right)$  
then finding the eigenvalues:  
$\lambda_{1,2}=2, \lambda_3=-1$  
and corresponding eigenvectors:  
$v_1=(-2,0,1), v_2=(-2,3,0), v_3=(-1,-1,1)$  
Here is where I am stuck. assuming I have done everything right so far(let me 
know if I have not), do I now just simply get each   
$g_i=e^{\lambda_it}v_i$  
and put them into equation 
$g(t)=C_1g_1+C_1g_2+C_3g_3$  
normally, or do I have to find another vector $u$ such that  
$(A-\lambda I)w=v_1\wedge g_2=e^{\lambda_2}(t\cdot v_1+u)$? 
If it is the second case, why do we need to do this? Why is the case of a duplicate eigenvalue special?
 A: Given the matrix
$$\begin{pmatrix}
 11 & 6 & 18 \\
 9 & 8 & 18 \\
 -9 & -6 & -16 \\
\end{pmatrix}$$
We find the two eigenvalues (one is repeated) using $|A - \lambda I| = 0$, and get
$$\lambda_{1,2} = 2, \lambda_3 = -1$$
Next, find the eigenvectors using $[A - \lambda_i I]v_i = 0$.
Using Gaussian Elimination with $\lambda_{1,2} = 2$, we get a Row-Reduced-Echelon-Form (RREF) of
$$\begin{pmatrix}
 1 & \dfrac{2}{3} & 2 \\
 0 & 0 & 0 \\
 0 & 0 & 0 \\
\end{pmatrix}v_{1,2} = 0$$
Because we have a repeated eigenvalue, we now try to find two linearly independent eigenvectors from the RREF result of $a = -\dfrac{2}{3} b - 2c$. Fortunately, we can get two linearly independent eigenvectors by choosing  $b = 0, c = 1$ and also $c = 0, b = 3$, giving eigenvectors 
$$v_1 = (-2, 0, 1)\\ v_2 = (-2, 3, 0)$$
As an aside, when we cannot find linearly independent eigenvectors, it is very important to learn the terms algebraic and geometric multiplicity, generalized eigenvectors and Jordan Form.
We repeat this process for the second eigenvalue $\lambda_3 = -1$ and find the third eigenvector
$$v_3 = (-1, -1 , 1)$$
As was mentioned in the comments by @amd, you have a complete eigenbasis for $A$, three linearly independent eigenvectors, so the matrix is diagonalizable. What do you know about the exponential of such a matrix?
Because of this, we can write the solution as
$$X(t) = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t} v_2 + c_3 e^{\lambda_3 t} v_3 = e^{2 t}\left(c_1 \begin{pmatrix}
 -2 \\ 0 \\ 1 \end{pmatrix} + c_2 \begin{pmatrix}
 -2 \\ 3 \\ 0 \end{pmatrix} \right) + c_3 e^{-t} \begin{pmatrix}
 -1 \\ -1 \\ 1 \end{pmatrix} $$
As was mentioned in the comments by @GitGud, that if we didn't have the two linearly independent eigenvectors, we would have had to find a generalized eigenvector or two and the solution would have been written as
$$X(t) =  c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t}(tv_1 + v_2) + c_3 e^{\lambda_3 t} v_3 $$
You can find more examples and details of what are called defective matrices,
generalized eigenvectors and Jordan Form here
 and here and here.
