Solving a system of differential equations using diagonalization 
Solve the system
  \begin{align*}
y'_1&= \phantom{-2}y_1 \\
y'_2&= -2y_1-4y_2
\end{align*}

I think they want me to solve it by using diagonalization. So far so good. I got the following:
The coefficient matrix for the system is
$$A=
\begin{bmatrix}
\phantom{-}1 & \phantom{-}0 \\
-2 & -4
\end{bmatrix}
$$
Then I find the eigenvalues
$$
\text{det}(A-\lambda I)=
\begin{bmatrix}
\phantom{-}1-\lambda & \phantom{-}0 \\
-2 & -4-\lambda
\end{bmatrix}=
(1-\lambda)(-4-\lambda)
$$
The eigenvalues are $\lambda=1$ and $\lambda=-4$.
Then I will find the matrix $P$ that $A$ is diagonalized by. $P$ is made up by the eigenvectors of $A$. So I find the eigenvectors.
For $\lambda=1$ 
$$
\begin{bmatrix}
\phantom{-}0 & \phantom{-}0 \\
-2 & -5
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=
\begin{bmatrix}
0 \\
0
\end{bmatrix}
$$
and for $\lambda=-4$
$$
\begin{bmatrix}
\phantom{-}5 & 0 \\
-2 & 0
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=
\begin{bmatrix}
0 \\
0
\end{bmatrix}
$$
Then I got stuck.  What are the eigenvectors?  I also need hints for the rest.
 A: i think you may not want to solve this equation by diagonalising it. this system is decoupled, therefore we can solve it directly. you have $y_1 = c_1e^t.$ subbing this in the second equation gives you, $$y_2' + 4y_2 = -2c_1e^t \tag 1$$ 
now $(1)$ has $y_2 = c_2e^{-4t}$ for a homogenous solution and $-\frac 25c_1e^t$ for a particular solution. therefore the solution to the system of equation is $$y_1 = c_1e^t, y_12 = c_2e^{-4t} - \frac 25c_1e^t.$$
A: To find an eigenvector corresponding to $\lambda = 1$, you need to determine values for $x_1$ and $x_2$ so that $0x_1+0x_2=0$ and $-2x_1-5x_2=0$.  One solution to this is $x_1=5, x_2=-2$, so you can use $(5, -2)$ as your eigenvector. 
You can perform a similar calculation to get an eigenvector corresponding to $\lambda=4$. Looking at $5x_1+0x_2=0$ and $-2x_1+0x_2=0$, we see that $(0,1)$ is an eigenvector. 
This means that $\vec{y_1}=(5,-2)e^{t}$ is a solution to your system, and so is $\vec{y_2}=(0,1)e^{-4t}$.  Since the two eigenvectors here are linearly independent, solutions to this system are linear combinations of $\vec{y_1}$ and $\vec{y_2}$.  Thus, the general solution to the system is:
$$\vec{y}=C_1(5,-2)e^t+C_2(0,1)e^{-4t},$$
where $C_1$ and $C_2$ are arbitrary constants.
