### Author Topic: Q6--T0101  (Read 5066 times)

#### Victor Ivrii

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##### Q6--T0101
« on: March 16, 2018, 08:08:07 PM »
a. Express the general solution of the given system of equations in terms of real-valued functions.
b. Also draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as $t\to \infty$.

$$\mathbf{x}' =\begin{pmatrix} 2 &-5\\ 1 &-2 \end{pmatrix}\mathbf{x}$$
« Last Edit: March 16, 2018, 08:10:36 PM by Victor Ivrii »

#### Cheng Sheng

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##### Re: Q6--T0101
« Reply #1 on: March 17, 2018, 01:51:27 AM »
The steps are shown below.

#### Victor Ivrii

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##### Re: Q6--T0101
« Reply #2 on: March 17, 2018, 05:11:01 AM »
See my comments to your another submission.

Why is your picture of that colour?!!

#### Junya Zhang

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##### Re: Q6--T0101
« Reply #3 on: March 18, 2018, 11:37:36 AM »
Cheng Sheng's solution is correct, but here's the typed solution
a)
Let $$P=\begin{pmatrix} 2 & -5 \\ 1 & -2\end{pmatrix}$$
Characteristic polynomial:
$$\chi_P(\lambda) = det\begin{pmatrix} 2-\lambda & -5 \\ 1 & -2-\lambda\end{pmatrix}=\lambda^2 + 1$$
Thus, $$\lambda_1 = i, \lambda_2 = -i$$
Consider $\lambda_1 = i$.
$$N(P-iI) = N\begin{pmatrix} 2-i & -5 \\ 1 & -2-i\end{pmatrix} = span\{\begin{pmatrix} 2+i \\ 1\end{pmatrix}\}$$
Consider $$e^{it}\begin{pmatrix} 2+i \\ 1\end{pmatrix} = \cos(t) + i\sin(t)\begin{pmatrix} 2+i \\ 1\end{pmatrix}\ = \begin{pmatrix} 2\cos(t) +2i\sin(t)+i\cos(t) - \sin(t) \\ \cos(t) + i\sin(t) \end{pmatrix} =\begin{pmatrix} 2\cos(t) - \sin(t) \\ \cos(t) \end{pmatrix} + i \begin{pmatrix} 2\sin(t) + \cos(t) \\ \sin(t)\end{pmatrix}$$
Thus, the general solution of this system is
$$\textbf{x}(t) = c_1\begin{pmatrix} 2\cos(t) - \sin(t) \\ \cos(t) \end{pmatrix} + c_2 \begin{pmatrix} 2\sin(t) + \cos(t) \\ \sin(t)\end{pmatrix}$$

b)
As $t\to\infty$, solution circulates in counter clockwise direction in elliptical shapes.
See attached image.