Find the general solution of each of the following systems (using method with constant coefficients).
I can complete most problems, but there are a few I cannot fully finish.
Problem One:
$\begin{cases} \frac{dx}{dt}=4x-2y \\ \frac{dy}{dt}=5x-2y \end{cases}\tag{1}$
$\left|\begin{pmatrix} 4-r & -2 \\ 5 & 2-r \end{pmatrix}\right| = (4-r)(2-r)+10 $ $(4-r)(2-r)+10 \iff r=3\pm3i$ \begin{equation} \begin{pmatrix} 4-r & -2 \\ 5 & 2-r\end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} \end{equation}
$=(4-r)A-2B$ $=5A+(2-r)B$
This is as far as I get without the problem becoming very messy..
Problem Two:
$\begin{cases} \frac{dx}{dt}=5x+4y \\ \frac{dy}{dt}=-x+y \end{cases}\tag{2}$
$\left|\begin{pmatrix} 5-r & 4 \\ -1 & 1-r \end{pmatrix} \right|=(5-r)(1-r)+4$ $r^2-6r+9=0 \iff r=3$
\begin{equation} \begin{pmatrix} 5-r & 4 \\ -1 & 1-r\end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} \end{equation}
$=(5-r)A+4B$ $=(-1)A+(1-r)B$
Now substitute for when $r=3$:
$2A+4B=0$ $-A-2B=0$
Using simple algebra we know $A=-2$ and $B=1$. Therefore, $x=-2e^{3t}$, and $y=e^{3t}$. Now we need a second solution of the form $x=(A_1+A_2t)e^t$, and $y=(B_1+B_2t)e^t$. Now I substitute these into our system of differential equation.
$(A_1+A_2t+A_2)e^t=5(A_1+A_2t)e^t+4(B_1+B_2t)e^t$ $(B_1+B_2t+B_2)e^t=-(A_1+A_2t)e^t+(B_1+B_2t)e^t$
Now, when I try finishing a problem like this one I get something different every time.
Could someone help me finish these two problems, so I can move on and finish the others that are similar to these. Thanks