Ode repeated roots I am looking for some help with the following question. I am having problems when I encounter repeated roots. For example 
$$
x' = \begin{bmatrix} 8& -1\\ 4& 12\end{bmatrix}x
$$
One solution is 
$$x^1 = \begin{bmatrix} 1\\- 2  \end{bmatrix}e^{10t}$$
But now I am searching for the second solution. the second solution comes in the form of 
$$x^2 = \begin{bmatrix} 1\\- 2  \end{bmatrix}te^{10t} + \begin{bmatrix} a\\b \end{bmatrix}e^{10t}$$
So my question is is there another way to determine the correct form for $x^2$ without just guessing and is there only one form that doesn't work and many others that do 
 A: Not quite an answer, but if you look at the Jordan form of the matrix
we have $J= \begin{bmatrix} 10 & 1 \\ 0 & 10 \end{bmatrix}$, and
so solving $y'=Jy$ will have $y_2(t) = y_2(0) e^{10t}$ and then solving the other will have
$y_1(t) = y_1(0) + y_2(0) \int_0^t e^{10 (t-\tau) }  e^{10 \tau}d \tau  = y_1(0) + y_2(0) t e^{10 t } $.
A: An alternative way is to take the Jordan decomposition of the matrix
$$
{\bf A} = \left( {\matrix{
   8 & { - 1}  \cr 
   4 & {12}  \cr 
 } } \right) = \left( {\matrix{
   0 & {1/4}  \cr 
   1 & {1/2}  \cr 
 } } \right)^{\, - \,{\bf 1}} \left( {\matrix{
   {10} & 1  \cr 
   0 & {10}  \cr 
 } } \right)\left( {\matrix{
   0 & {1/4}  \cr 
   1 & {1/2}  \cr 
 } } \right) = {\bf C}^{\, - \,{\bf 1}} \;{\bf D}\;{\bf C}
$$
So you get
$$
{\bf x}' = {\bf A}\,{\bf x}\quad  \Rightarrow \quad \left( {{\bf C}\;{\bf x}} \right)' = \;{\bf D}\;\left( {{\bf C}\,{\bf x}} \right)\quad  \Rightarrow \quad {\bf y}' = \;{\bf D}\;{\bf y}
$$
and
$$
\left\{ \matrix{
  y_1 ' = 10y_1  + y_2  \hfill \cr 
  y_2 ' = 10y_2  \hfill \cr}  \right.\quad  \Rightarrow \quad \left\{ \matrix{
  y_1 ' = 10y_1  + c_2 \,e^{10\,t}  \hfill \cr 
  y_2  = c_2 \,e^{10\,t}  \hfill \cr}  \right.\quad  \Rightarrow \quad \left\{ \matrix{
  y_1  = c_1 \,e^{10\,t}  + c_2 t\,e^{10\,t}  \hfill \cr 
  y_2  = c_2 \,e^{10\,t}  \hfill \cr}  \right.
$$
and finally put
$$
{\bf x} = {\bf C}^{\, - \,{\bf 1}} \,{\bf y}
$$
A: Here's another way:
Let $A$ be the given coefficient matrix:
$A = \begin{bmatrix} 8 & -1 \\ 4 & 12 \end{bmatrix}; \tag 1$
then the characteristic polynomial of $A$ is
$p_A(x) = \det(A - xI) = x^2 - \text{Tr}(A) + \det A; \tag 2$
we have
$\text{Tr}(A) = 20, \; \det A = 100, \tag 3$
so
$p_A(x) = x^2 - 20x + 100 = (x - 10)^2; \tag 4$
it follows from (4) that $A$ has one eigenvalue $10$ of multiplicity $2$, and from this it further follows that $A$ can not be diagonalized; for if $A$ could be diagonalized, there would exist a non-singular $2 \times 2$ matrix $P$ such that
$10 I = \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix} = PAP^{-1}, \tag 5$
and (5) implies
$A = P^{-1}(10I)P = 10(P^{-1} I P) = 10I; \tag 6$
this contradicts (1) and so shows $A$ is non-diagonalizable.  Now set
$N = A - 10I = \begin{bmatrix} -2 & -1 \\ 4 & 2 \end{bmatrix}; \tag 7$
we see that
$N^2 = 0. \tag 8$
Now we recall that any solution of 
$x' = Ax \tag 9$
with
$x(t_0) = x_0 \tag {10}$
is of the form
$x(t) = e^{A(t - t_0)} x_0; \tag{11}$
from (7) we have
$A = 10I + N, \tag{12}$
so
$x(t) = e^{A(t - t_0)}x_0 = e^{(10I + N)(t - t_0)}x_0 = e^{10I(t - t_0) + N(t - t_0)}x_0; \tag{13}$
since $IN = NI$ we may write
$e^{10I(t - t_0) + N(t - t_0)} = e^{10I(t - t_0)}e^{N(t - t_0)}; \tag{14}$
now
$e^{10I(t - t_0)} = e^{10(t - t_0)}I, \tag{15}$
as is easily seen.  As for $e^{N(t - t_0)}$, we use (8) in the power series of matrix exponential and find
$e^{N(t - t_0)} = I + (t - t_0)N, \tag{16}$
since the terms in $N^2$ and higher powers vanish.  We bring it all together using (13)-(16):
$x(t) = e^{10(t - t_0)}(I + (t - t_0)N)x_0. \tag{17}$
If $x_0 \ne 0$ is such that $Nx_0 = 0$, for example if
$x_0 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}, \tag{18}$
then the solution $x(t)$ only involves terms in $e^{10(t -t_0)}$; otherwise, with $Nx_0 \ne 0$, $x(t)$ exhibits a functional dependence on both $e^{10(t - t_0)}$ and $(t - t_0)e^{10(t - t_0)}$, and we see from the above derivations, as well as from the standard results for uniqueness of solutions to ODEs, that there are no other solutions possible.  
In writing this answer, I have been able to avoid explicit invocation of the Jordan form via knowledge of the fact that $N^2 = (A - 10I)^2 = 0$, a result which may be seen from the Jordan form of $A$ (see the answers of G Cab and copper.hat) or in other ways.  I like the method given here because it avoids a lot of the computation required to reduce $A$ to cannonical form.
