For the matrix $A = \begin{pmatrix}-4 & 2 \\ 2 & -1 \end{pmatrix}$, I need to find the real, general solution of the matrix equation $$Y^{\prime} = AY, $$ where $Y = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}$.
To start, I found that the eigenvalues of $A$ are $r_{1} = 0$ and $r_{2} = -5$, and that the eigenvectors corresponding to $r_{1} = 0$ are $u_{1} = s\begin{pmatrix}1 \\ 2 \end{pmatrix}$ and $u_{2} = s \begin{pmatrix} -2 \\ 1 \end{pmatrix}$.
Now, taking $s=1$ in each of these gives us the corresponding eigenvectors $u_{1} = \begin{pmatrix}1 \\ 2 \end{pmatrix}$ and $u_{2} = \begin{pmatrix}-2 \\ 1 \end{pmatrix}$, which I checked were linearly independent.
Therefore, a general solution should be $Y(t) = c_{1}e^{0\cdot t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_{2} e^{-5t}\begin{pmatrix} -2 \\ 1 \end{pmatrix} = c_{1} \begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_{2}e^{-5t}\begin{pmatrix} -2 \\ 1 \end{pmatrix}$, or
$$x(t) = c_{1}-2c_{2}e^{-5t} \\ y(t) = 2c_{1} + c_{2}e^{-5t} $$
To check, I differentiated them and plugged $x(t)$ and $y(t)$ back into $AY$, and wound up getting $Y^{\prime}$ back, but I still wanted an extra set of eyes to tell me what I did was okay.
Also, in the case where one of my eigenvalues is $0$, like it is here, is there anything special I need to do, like multiply by an extra $t$ or something? I don't know - it just seems weird. (I'm kind of teaching myself this, so forgive me if I ask inane questions.)
Thank you in advance.