System of differential equations verification I have a two questions about these types of equations:


*

*Is on checking if I had solve one case correctly

*A simpler way to solve/represent the solutions of it by using Jordan forms I think
So I want to solve this system

$x' = -23x -8y$
$y' = 60x +21y$

I first find eigen values with
$
det
  \begin{bmatrix}
    -23-\lambda & -8 \\
    60 & 21-\lambda
  \end{bmatrix} =0 \therefore \lambda_1 =1, \lambda_2=-3
$
I then find the eigen vectors $v$(with $\lambda_1 $) and $u$(with $\lambda_2 $) using the each eigen value by solving
$
\begin{pmatrix}
    -24 & -8 \\
    60 & 20
\end{pmatrix}\begin{pmatrix}
    v_1 \\
    v_2
\end{pmatrix} =\begin{pmatrix}
    0 \\
    0
\end{pmatrix} \therefore 3v_1=-v_2
$
$
\begin{pmatrix}
    -20 & -8 \\
    60 & 24
\end{pmatrix}\begin{pmatrix}
    u_1 \\
    u_2
\end{pmatrix} =\begin{pmatrix}
    0 \\
    0
\end{pmatrix} \therefore u_2=\frac{-5u_1}{2}
$
then we choose $v_1 = 1$ and $u_1=2$, arbitrarily:
$
\begin{pmatrix}
    v_1 \\
    v_2
\end{pmatrix} =\begin{pmatrix}
    1 \\
    -3
\end{pmatrix}
$
$
\begin{pmatrix}
    u_1 \\
    u_2
\end{pmatrix} =\begin{pmatrix}
    2 \\
    -5
\end{pmatrix}
$
then the general solutions are $ y(t)=Ce^{Dt}C^{-1}y_0 $ :
$
y(t)= \begin{pmatrix}
    1&2 \\
    -3&-5
\end{pmatrix} \begin{pmatrix}
    e^{\lambda_1 t}&0 \\
    0&e^{\lambda_2 t}
\end{pmatrix} \begin{pmatrix}
    -5&-2 \\
    3&1
\end{pmatrix} \begin{pmatrix}
    x_0\\
    y_0
\end{pmatrix}, \lambda_1 = 1, \lambda_2 =-3
$
How might I then find a specific solution given initial conditions?
In class the teacher showed a simpler way to represent the solutions with an example below, how may I apply that to the this question(above)?

$x' =-3x+y$
$y' =-3y$

$
\begin{pmatrix}
    x' \\
    y'
\end{pmatrix} = \begin{pmatrix}
    -3&1 \\
    0&-3
\end{pmatrix}\begin{pmatrix}
    x \\
    y
\end{pmatrix}
$
and then they said something about $\begin{pmatrix}
    -3&1 \\
    0&-3
\end{pmatrix}$ being in Jordan form(for 2 by 2 matrix) and so $\lambda =-3, k_{dimension-of- matrix}=2 $ then:
$
\begin{pmatrix}
    x \\
    y
\end{pmatrix} = e^{-3t}\begin{pmatrix}
    1&t \\
    0&1
\end{pmatrix}\begin{pmatrix}
    x_0 \\
    y_0
\end{pmatrix} 
$

$$y=c_1e^{\lambda _1t }v_1 +c_2 e^{\lambda _2t }v_2=$$
$$ c_1e^{t } \begin{pmatrix}
    1 \\
    -3
\end{pmatrix}  +c_2 e^{2t }    \begin{pmatrix}
    2 \\
    -5  \end{pmatrix}  $$
You can find $c_1$ and $c_2$ using a given initial value.
