Particular solution to system of differential equations (with eigen- vectors & values) Given a system of differential equations
$y_1'=2y_1+y_2$
$y_2'=-2y_1+5y_2$
Which has the eigenvalues 3 and 4 for the total matrix of the above mentioned system. Where the vectors $(1,1)$ and $(1,2)$ are eigenvectors to the above mentioned eigenvalues. Such that $(1,1)$ is the eigenvector of the value 3 and the other vector of 4.
1) Find the solution for the differential system
2) Find the particular solution such that
$$\lim_{t\to \infty}   e^{-t4}y_1(t)=0$$
and $$y_2'(0)=3$$
1) I found the solution 
$c_1e^{3t}(1,1)+c_1e^{4t}(1,2)$
2) I'm uncertain with the second part. The limit creates some issues for me. 
 A: Write the system in matrix form, $y' = Ay$. Note that if an initial condition $y_0$ is an eigenvector of $A$ corresponding to eigenvalue $\lambda$ and we
set $y(t) = e^{\lambda t} y_0$, then
$y'(t) = \lambda e^{\lambda t} y_0 = e^{\lambda t} A y_0 = A (e^{\lambda t} y_0) = A y(t)$, so $y(t) = e^{\lambda t} y_0$ is a solution.
You are given two eigenvectors $v_1,v_2$ corresponding to eigenvalues $\lambda_1, \lambda_2$, so you have two solutions of the form above (in fact,
since the system is linear, any linear combination is also a solution, and since
there are two of them and the are linearly independent, the linear combinations form all of the solutions).
Now pick an eigenvector $v$ corresponding to an eigenvalue $\lambda$ so that
$e^{-4t} e^{\lambda t} v \to 0$.
A: We have
$$
A =
\begin{pmatrix}
2 & 1 \\
-2 & 5
\end{pmatrix}
$$
and
$$
y' = A y
$$
Then
$$
A (1,1)^T = (3,3)^T = 3 (1,1)^T \\
A (1,2)^T = (4,8)^T = 4 (1,2)^T
$$
The solution of the linear ODE is:
$$
0 = y' - A y = (e^{-tA}y)' \Rightarrow \\
y(t) = e^{tA} y(0)
$$
Further for an eigenvector $y_\lambda$ with eigenvalue $\lambda$ we have
$$
e^{tA} y_\lambda
=\sum_{k=0}^\infty \frac{1}{k!}t^k A^k y_\lambda
=\sum_{k=0}^\infty \frac{1}{k!}t^k \lambda^k y_\lambda
= e^{\lambda t} y_\lambda
$$
So here we get
$$
y(t) = c_1 e^{3t}(1,1)^T + c_2 e^{4t}(1,2)^T
$$
as general solution. This corresponds with your solution.
Applying the first condition to the general solution yields
\begin{align}
0 
&= \lim_{t\to\infty} e^{-4t} y_1(t) \\
&= \lim_{t\to\infty} e^{-4t} (c_1 e^{3t} + c_2 e^{4t}) \\
&= c_1 \left( \lim_{t\to\infty} e^{-4t} e^{3t} \right) + 
   c_2 \left(\lim_{t\to\infty} e^{-4t} e^{4t} \right)) \\
&= c_1 \left( \lim_{t\to\infty} e^{-t} \right) + 
   c_2 \left( \lim_{t\to\infty} 1 \right)\\
&= c_2
\end{align}
and further
$$
3 
= y_2'(0) 
= \left. (3 c_1 e^{3t} + 8 c_2 e^{4t}) \right\vert_{t=0}
= 3 c_1 + 8 c_2
= 3 c_1 
$$
So we get $c_1 = 1$ and $c_2 = 0$ as coefficients of the particular solution:
$$
y_p(t) =  e^{3t}(1,1)^T
$$
