MAT244--2019F > Term Test 2

Problem 4 (morning)

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Victor Ivrii:
Find the general real solution to
$$\mathbf{x}'=\begin{pmatrix} 2 & -3\\ 4 &-2\end{pmatrix}\mathbf{x}$$
and sketch trajectories.

Yiheng Bian:
$$det(A-{\lambda}I)=0\\ \begin{vmatrix} 2-\lambda & -3 \\ 4 & -2-\lambda \end{vmatrix}={\lambda}^2+8=0$$
So
$$\lambda_1=\sqrt{8}i\\ \lambda_2=-\sqrt{8}i$$
$$\text{when } \lambda=\sqrt{8}I\\ \begin{vmatrix} 2-\sqrt{8}i & -3 \\ 4 & -2-\sqrt{8}i \end{vmatrix} = \begin{vmatrix} 0 \\ 0 \end{vmatrix}$$
RREF:
$$\begin{pmatrix} 2-\sqrt{8}i & -3 \\ 0 & 0 \end{pmatrix} \quad = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad$$
Let x_2=t
So we can get:
$${(2-\sqrt{8}i})x_1=3x_2=3t\\ x_1=\frac{3t}{2-\sqrt{8}i}$$
So
$$t*\begin{pmatrix} \frac{3}{2-\sqrt8i} \\ 1 \end{pmatrix} \quad$$
Therefore:
$$e^{i\sqrt8t}\begin{pmatrix} \frac{3}{2-\sqrt8i} \\ 1 \end{pmatrix} \quad = (cos(\sqrt8t)+isin(\sqrt8t))\begin{pmatrix} \frac{3}{2-\sqrt8i} \\ 1 \end{pmatrix} \quad=\begin{pmatrix} \frac{cos(\sqrt8t)-\sqrt2sin(\sqrt8t)}{2} +\frac{isin(\sqrt8t)+i\sqrt2cos(\sqrt8t)}{2}\\ cos(\sqrt8t )+ isin(\sqrt8t)) \end{pmatrix} \quad$$
So
$$y=c_1\begin{pmatrix} \frac{cos(\sqrt8t)-\sqrt2sin(\sqrt8t)}{2} \\ cos(\sqrt8t) \end{pmatrix} \quad +c_2\begin{pmatrix} \frac{sin(\sqrt8t)+\sqrt2cos(\sqrt8t)}{2}\\ sin(\sqrt8t)) \end{pmatrix} \quad$$
OK, except LaTeX sucks:

1) * IS NOT a sign of multiplication
2)  "operators" should be escaped: \cos, \sin, \tan, \ln

Lan Cheng:
a) First we should solve for eigenvalues:

let $det\begin{bmatrix}2-\lambda & -3\\ 4 & -2-\lambda \end{bmatrix}=0.\lambda_{1}=2\sqrt{2}i,\lambda_{2}=-2\sqrt{2}i.$

Second we need to solve for eigenvectors:

Let $(A-\lambda I)v=0.$ When $\lambda=2\sqrt{2}i,v=\begin{bmatrix}\sqrt{2}i+1\\ 2 \end{bmatrix}.$

Therefore, $e^{(2\sqrt{2}i)t}\begin{bmatrix}\sqrt{2}i+1\\ 2 \end{bmatrix}=(cos(2\sqrt{2}t)+isin(2\sqrt{2}t))\begin{bmatrix}\sqrt{2}i+1\\ 2 \end{bmatrix}.$

$x(t)=C_{1}\begin{bmatrix}-\sqrt{2}sin(2\sqrt{2}t)+cos(2\sqrt{2}t)\\ 2cos(2\sqrt{2}t) \end{bmatrix}+C_{2}\begin{bmatrix}\sqrt{2}cos(2\sqrt{2}t)+sin(2\sqrt{2}t)\\ 2sin(2\sqrt{2}t) \end{bmatrix}.$

b) See photo below.

Ruojing Chen:
$$x'= \left ( \begin{matrix} 2 & -3 \\ 4 & -2 \end{matrix} \right ) x$$

$$det(A-\lambda I)=0$$
$$(2-\lambda)(-2-\lambda)-(-3)*4=0$$
$$-4+\lambda^2+12=0$$
$$\lambda^2=-8$$
$$\Rightarrow \lambda=\pm2\sqrt{2}i$$

$$when \lambda=2\sqrt{2}i$$
$$\left ( \begin{matrix} 2-2\sqrt{2}i & -3 \\ 4 & -2-2\sqrt{2}i \end{matrix} \right ) \left ( \begin{matrix} x_1 \\ x_2 \end{matrix} \right ) =\left ( \begin{matrix} 0\\ 0 \end{matrix} \right )$$
$$(2-2\sqrt{2}i)x_1=3x_2$$
$$\Rightarrow x=t \left ( \begin{matrix} 3 \\ 2-2\sqrt{2}i \end{matrix} \right )$$

$$e^{2\sqrt{2}it}=t \left ( \begin{matrix} 3 \\ 2-2\sqrt{2}i \end{matrix} \right )$$
$$=(cos2\sqrt{2}t+isin2\sqrt{2}t)\left ( \begin{matrix} 3 \\ 2-2\sqrt{2}i \end{matrix} \right )$$

$$\therefore =c_1 \left( \begin{matrix} 3cos2\sqrt{2} \\ 2cos2\sqrt{2}+2\sqrt{2}sin2\sqrt{2} \end{matrix} \right )+ c_2\left(\begin{matrix} 3sin2\sqrt{2} \\ 2sin2\sqrt{2}-2cos2\sqrt{2} \end{matrix} \right )$$

Ruojing Chen:
a) sketch