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.