Solving a Differential equation using eigenvectors I want to solve the following differential equation using eigenvectors:
$x'=y$
$y' = 2x-5y$
Or using matrices: 
$\begin{pmatrix} 
x' \\
y' 
\end{pmatrix} = \begin{pmatrix} 
0 & 1 \\
2 & -5 
\end{pmatrix} \begin{pmatrix} 
x \\
y 
\end{pmatrix}$
With the initial conditions $x(0) = 0$ and   $y(0) = 2$
Now I know that all I'd have to do is essentially find the exponential of the matrix
 $$ A =\begin{pmatrix}0 & 1 \\2 & -5 \end{pmatrix}$$ but my teacher wants us to solve it using eigenvectors.
Let's find A's eigenvectors:
$det(A-\lambda I) = -\lambda(-5-\lambda)-2 = \lambda^2+5\lambda-2$
A's eigenvalues are:


*

*$\lambda_1 = \dfrac{-5+\sqrt{33}}{2}$ 

*$\lambda_2 = \dfrac{-5-\sqrt{33}}{2}$
Now what do I do to solve the differential equation using A's eigenvector?

You have done most of the work already. $A's$ eigenvalues are:
$$\lambda_{1, 2} = \dfrac{-5\pm\sqrt{33}}{2}$$ 
We now have to find the eigenvectors and solve $[A - \lambda_i I]v_i = 0$. So for $\lambda_{1,2}$, we have:
$$[A - \lambda_{1,2} I]v_{1,2} = [A - \dfrac{1}{2} (-5 \pm \sqrt{33}) I]v_{1,2} = \begin{pmatrix} 
- \dfrac{1}{2} (-5 \pm \sqrt{33}) & 1 \\
2 & -5 - \dfrac{1}{2}(-5 \pm \sqrt{33})
\end{pmatrix}v_{1, 2} = \begin{pmatrix} 
0 \\
0
\end{pmatrix}$$
When we do the RREF of this matrix, we have:
$$\begin{pmatrix} 
1 & \dfrac{1}{4} (-5 \mp\sqrt{33}) \\
0 & 0 \end{pmatrix}v_{1, 2}= \begin{pmatrix} 
0 \\
0
\end{pmatrix}$$
This gives us:
$$v_{1,2} = \begin{pmatrix} 
 \dfrac{1}{4}(5 \pm \sqrt{33})\\1
\end{pmatrix}$$
We can now write:
$$X(t) = \begin{pmatrix} 
 x(t)\\y(t)
\end{pmatrix} = c_1 e^{\lambda_1 t}  v_1 + c_2 e^{\lambda_2 t} v_2$$
You have two initial conditions to solve for $c_1$ and $c_2$ and should get:
$$X(t) = \begin{pmatrix} 
 x(t)\\y(t)
\end{pmatrix} =\begin{pmatrix} 
 \dfrac{2}{\sqrt{33}}~e^{-1/2 (5+\sqrt{33}) t} (e^{\sqrt{33} t}-1) \\ \dfrac{1}{33} e^{-1/2 (5+\sqrt{33}) t} ((33-5 \sqrt{33}) e^{\sqrt{33} t}+33+5 \sqrt{33})
\end{pmatrix}$$
Update For $\lambda_1$, our RREF is:
$$\begin{pmatrix} 
1 & \dfrac{1}{4} (-5 -\sqrt{33}) \\
0 & 0 \end{pmatrix}v_1= \begin{pmatrix} 
0 \\
0
\end{pmatrix}$$
Let the vector $v_1 = \begin{pmatrix} a \\ b \end{pmatrix}$, so this gives us:
$$a + \dfrac{1}{4} (-5 -\sqrt{33})b = 0 \implies a = \dfrac{1}{4} (5 +\sqrt{33})b$$
We are free to choose whatever $b$ we'd like, so choose $b = 1$ and conclude.
This is how we arrived at:
$$v_1 = \begin{pmatrix} 
 \dfrac{1}{4}(5 + \sqrt{33})\\1
\end{pmatrix}$$
I should also mention, note how this choice also satisfies the second equation.
