Toronto Math Forum
MAT244--2019F => MAT244--Test & Quizzes => Term Test 2 => Topic started by: Victor Ivrii on November 19, 2019, 04:23:46 AM
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Find the general real solution to
$$
\mathbf{x}'=\begin{pmatrix}
2 & -3\\
4 &-2\end{pmatrix}\mathbf{x}
$$
and sketch trajectories.
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$$
det(A-{\lambda}I)=0\\
\begin{vmatrix}
2-\lambda & -3 \\
4 & -2-\lambda
\end{vmatrix}={\lambda}^2+8=0
$$
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
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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.
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$$x'= \left (
\begin{matrix}
2 & -3 \\
4 & -2
\end{matrix}
\right ) x$$
$$det(A-\lambda I)=0$$
$$(2-\lambda)(-2-\lambda)-(-3)*4=0$$
$$-4+\lambda^2+12=0$$
$$\lambda^2=-8$$
$$\Rightarrow \lambda=\pm2\sqrt{2}i$$
$$when \lambda=2\sqrt{2}i$$
$$\left (
\begin{matrix}
2-2\sqrt{2}i & -3 \\
4 & -2-2\sqrt{2}i
\end{matrix}
\right ) \left (
\begin{matrix}
x_1 \\
x_2
\end{matrix}
\right ) =\left (
\begin{matrix}
0\\
0
\end{matrix}
\right ) $$
$$(2-2\sqrt{2}i)x_1=3x_2$$
$$\Rightarrow x=t
\left (
\begin{matrix}
3 \\
2-2\sqrt{2}i
\end{matrix}
\right )$$
$$e^{2\sqrt{2}it}=t
\left (
\begin{matrix}
3 \\
2-2\sqrt{2}i
\end{matrix}
\right )$$
$$=(cos2\sqrt{2}t+isin2\sqrt{2}t)\left (
\begin{matrix}
3 \\
2-2\sqrt{2}i
\end{matrix}
\right )$$
$$\therefore =c_1
\left(
\begin{matrix}
3cos2\sqrt{2} \\
2cos2\sqrt{2}+2\sqrt{2}sin2\sqrt{2}
\end{matrix}
\right )+ c_2\left(\begin{matrix}
3sin2\sqrt{2} \\
2sin2\sqrt{2}-2cos2\sqrt{2}
\end{matrix}
\right )$$
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a) sketch
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I found the general real solution.
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This is the solution
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:)
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What everybody is missing:
we see that characteristic roots $k_{1,2}= \pm \sqrt{8}i$ are purely imaginary. So, it is center and with counter-clock-wise orientation since the bottom-left element is positive.