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