Question:
a) Find the general solution of the given system of equations.
b) Draw a directions field and a few of the trajectories. In each of these problems, the coefficient matrix has a zero eigenvalue. As a result, the pattern of trajectories is different from those in the examples in the text.
\begin{align*}
x' =
\begin{pmatrix}
3 & 6 \\
-1 & -2 \\
\end{pmatrix}
x \\
\end{align*}
Solution:
\begin{align*}
det
\begin{pmatrix}
3-\lambda & 6 \\
-1 & -2-\lambda \\
\end{pmatrix}
&=0 \\
(3-\lambda)(-2-\lambda)-(-6) &=0 \\
\lambda^2-\lambda &=0 \\
\lambda(\lambda-1) &=0 \\
\lambda = 0 \quad or \quad \lambda = 1 \\
\end{align*}
Case 2: $\lambda$=0
\begin{align*}
\begin{pmatrix}
3 & 6 & \vdots & 0\\
-1 & -2 & \vdots & 0\\
\end{pmatrix}
\quad \sim \quad
\begin{pmatrix}
3 & 6 & \vdots & 0\\
0 & 0 & \vdots & 0\\
\end{pmatrix}
\quad \sim \quad
\begin{pmatrix}
\frac{1}{2} & 1 & \vdots & 0\\
0 & 0 & \vdots & 0\\
\end{pmatrix} \\
\end{align*}
\begin{align*}
\Rightarrow \qquad
\begin{pmatrix}
x1 \\ x2
\end{pmatrix}
=
\begin{pmatrix}
-2 \\ 1
\end{pmatrix}
t
\end{align*}
Case 1: $\lambda$=1
\begin{align*}
\begin{pmatrix}
3-1 & 6 & \vdots & 0\\
-1 & -2-1 & \vdots & 0\\
\end{pmatrix}
\quad \sim \quad
\begin{pmatrix}
2 & 6 & \vdots & 0\\
-1 & -3 & \vdots & 0\\
\end{pmatrix}
\quad \sim \quad
\begin{pmatrix}
\frac{1}{3} & 1 & \vdots & 0\\
0 & 0 & \vdots & 0\\
\end{pmatrix}
\end{align*}
\begin{align*}
\Rightarrow \qquad
\begin{pmatrix}
x1 \\ x2
\end{pmatrix}
=
\begin{pmatrix}
-3 \\ 1
\end{pmatrix}
t
\end{align*}
Therefore, the general solution is
\begin{align*}
y = c1 *
\begin{pmatrix}
-2 \\ 1
\end{pmatrix}
+ c2 * e^t
\begin{pmatrix}
-3 \\ 1
\end{pmatrix}
\end{align*}
