Matrix ODE, defective eigenvalue: Where does the extra '$t$' come from? 
Given $A \in \Bbb R^{2\times 2}$, the system
  $$
\dot X=AX
$$
Has the solution 
  $$
X= c_1e^{\lambda t}\xi_1+c_2e^{\lambda t}\left (\xi_1 t+\xi_2 \right)
$$
Where $\xi_1$ is the unique eigenvector associated to the double eigenvalue $\lambda$ and $\xi_2$ satisfies:
$$
(A-\lambda I)\xi_2=\xi_1
$$

I've seen a proof of this fact, but I still don't understand why the second linearly independent solution has that form: Where does the multiplying $t$ come from?
I'm interested in a linear algebra perspective of this.
 A: I think that linear algebra perspective is the simplest way to explain where does this additional $t$ come from. As you know, you can always make a substitution that transforms system 
$$\dot{x}=Ax$$ 
into 
$$\dot{y}=Jy,$$
 where $J$ is a Jordan normal form of matrix $A$. Also you might know that the solution to Cauchy problem 
$$\dot{x}=Ax, \; x(0)=x_0$$ 
is given by formula 
$$x(t) = \exp At \cdot x_0.$$ 
If you have a block diagonal matrix $A = {\rm diag}\, (A_1, A_2, \dots, A_n)$, then $\exp A$ is again block diagonal and 
$$\exp A = (\exp A_1, \exp A_2, \dots, \exp A_n).$$ 
Combining all these facts, we can focus our attention on what is the result of matrix exponentiation applied to Jordan block. Each Jordan block $J_i$ could be written as $D_i + N_i$, where $D_i$ is the diagonal part of matrix $J_i$ and $N_i$ is nilpotent part of the same matrix. These two matrices commute, and there is another property of exponent which states that if $$\lbrack A, B \rbrack = 0$$ then $$\exp (A + B) = \exp A \cdot \exp B.$$ Also, note that if $N_i \in \mathbb{R}^{k \times k}$, then $N^k_i = 0$. The exponent of scalar matrix $D_i t = d_i t \cdot E$ is just a $\exp(d_i t) \cdot E$; multiplying by this matrix is the same as multiplying by the scalar function $e^{d_i t}$. Returning back to the nilpotent part, we observe that power series for $\exp N_i t$ truncates after term $N^{k-1}_i$ and gives us just a matrix whose entries are polynomial in $t$. Well, this is how polynomial terms appear in general solution of system of linear equations.
