Find the general solution of each of the following systems (using method with constant coefficients). 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
 A: For the first problem, there's no way around working with "messy" computations.
When you write:
$$= (4-r)A - 2B$$
$$= 5A +(2-r)B$$
...what exactly is each expression equal to? Are you simply using '$\;=\;$' to denote first and second entries, respectively, of a $2\times1$ column matrix?
Note that in your second problem there is a problem with the following:

Now substitute for when $r=3$:
$$2A+4B=0$$
$$-A-2B=0$$
Using simple algebra we know $A=-2$ and $B=1$.

If you attempt to solve the system of equations for A and B, if you multiply $(-A-2B=0)$ by $2$ you get $-2A - 4B =0$. Now, adding that equation to the first gives you $0=0$, and hence, there are an infinite number of solutions, $(A, B)$. (The equations are linearly dependent.)
A: You also can solve the system $$\begin{cases} \frac{dx}{dt}=4x-2y \\ \frac{dy}{dt}=5x-2y \end{cases}
$$ as follows. Let $D$ stands for differentiation, so the system will be: $$\begin{cases} Dx=4x-2y \\ Dy=5x-2y \end{cases}
\longrightarrow\begin{cases} (D-4)x+2y=0 \\ -5x+(D+2)y=0 \end{cases}$$ Now solve this system by using very elementary way. For example, I multiply the first equation by $5$ and the second by $(D-4)$ and then adding the results, we have:
$$(D^2-2D+2)y=0$$ By this way we eliminated the variable $x$. The latter ODE is equivalent to $$y''-2y'+2y=0$$ which can be easily solved. Indeed, we get $$y(t)=C_1e^t\sin t+C_2e^t\cos t$$ Now do the same way for finding an ODE with respect to $x$ and then finding $x(t)$.
