Linear ODE repeated eigenvalues how to find more than 2 generalized eigenvectors So I've searched around the web for a few hours now, as
(i)
$\mathbf A = \begin{pmatrix}2&1\\0&2\end{pmatrix}$
The characteristic polynomial is $(\lambda-2)^2=0$, so $\lambda=2$, repeated.
A bit of algebra convinces us that $b=(1,0)^t$ is an eigenvector.
Setting $(\mathbf A-\lambda \mathbf I)r=b$, with $r$ and $b$ vectors. Maybe a bit unmotivated but lets proceed. We find that $r=(0,1)^t$ is a solution. Now the solution is 
$x(t)= c_1*e^{2t}+c_2(te^{2t}b+e^{2t}r)$
However lets say I have 3 or more repeated eigenvalues, or two eigenvalues that are both eigenvalues what then?
For instance: 
$\mathbf A = \begin{pmatrix}1&2&0\\0&1&3/4\\0&0&1\end{pmatrix}$
I find $\lambda=1$ and the vector $b_2=(1,0,0)^t$ is a solution. But I'm not sure how to proceed. 
Any help would be appreciated.

Write $A = \lambda I + N$ where $N$ is nilpotent (in your example, $N = \pmatrix{0 & 2 & 0\cr 0 & 0 & 3/4\cr 0 & 0 & 0\cr}$ with $N^3 = 0$).  Suppose $N^k = 0$ but $N^{k-1} \ne 0$.  Then your ODE $y' = A y$ has solutions of the form $y(t) = \sum_{j=0}^{k-1} \frac{t^j}{j!} e^{\lambda t} N^j u$ for any vector $u$.  Take a suitable basis...
