Solve $\textbf y'=A\textbf y$ with $\textbf y\in \mathbb R^4$ and $A\in \text{Mat}(4\times 4,\mathbb R)$ 
We consider
  $$\textbf y'(t)=A\textbf y(t)$$
  with $\textbf y(0)=\textbf y_0\in \mathbb R^4$ and $A\in \text{Mat}(4\times 4,\mathbb R)$. Let $\textbf y_1,\textbf y_2,\textbf y_3,\textbf y_4\in\mathbb R^4$ linearly independant and $\lambda_1\neq \lambda_2\neq \lambda_3\in\mathbb C$ s.t.
  $$A\textbf y_1=\lambda_1\textbf y_1$$
  $$A\textbf y_2=\lambda_2\textbf y_2$$
  $$A\textbf y_3=\lambda_3\textbf y_3$$
  $$(A-\lambda_3)\textbf y_4=\textbf y_3.$$
Write the solution $\textbf y(t)$ in function of $\textbf y_0, \textbf y_i$ and $\lambda_i$. Give a condition such that $$\limsup_{t\to\infty }\textbf y(t)<\infty .$$

I agree that $\textbf y_1,\textbf y_2,\textbf y_3$ are eigenvectors and $\lambda_1,\lambda_2,\lambda_3$ eigenvalues, but what can I do with $\textbf y_4$ ? Because I can not diagonalize this matrix. And even if I could, I would like to calculate the $e^A$, but it doesn't look easy since I don't have the change of basis. It's probably not a complicate exercise, but like that, I can't continue. 
 A: Note that for $y_4$ you have
$$
(A-\lambda_3I)^2 y_4=0.
$$
You can use this fact and the definition of the matrix exponent to find that
$$
y(t)=e^{\lambda_3 t}(I+(A-\lambda_3I)t)y_4
$$
solve your system. Now you have four linearly independent solutions and hence can write down the general solution.
Here are some additional details.
The task is to calculate $e^{tA}y_4$. Note that you do not need to know $e^{tA}$ to actually calculate the needed expression. 
Start with $$e^{tA}y_4=e^{\lambda_3tI-\lambda_3tI+tA}=e^{\lambda_3 t}e^{t(A-\lambda_3I)}y_4,$$
and finish the calculations using the definition.
A: here is a nother way to do this. let $y= a_1y_1 + a_2y_2 + a_3y_3 + a_4y_4,$ then $y^\prime = Ay$ turns into $a_1^\prime = \lambda_1 a_1, a_2^\prime = \lambda_2 a_2, a_3^\prime = \lambda_3 + a_4, a_4^\prime = \lambda_3 y_4.$ the solutions are
$$a_1 = a_1(0)e^{\lambda_1t},
 a_2 = a_2(0)e^{\lambda_2t},
 a_3 = (a_3(0) + ta_4(0))e^{\lambda_3t},
 a_4 = a_4(0)e^{\lambda_3t}$$ 
therefore the solution is $$ y = ae^{\lambda_1t}+
 be^{\lambda_2t}+
 (c + d + td)e^{\lambda_3t} \text { where }a, b, c, d \text{ are constants.}$$
the condition for existence for all $t> 0$ is that all eigenvalues be negative.
