Author Topic: Q6--T5101  (Read 2226 times)

Victor Ivrii

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Q6--T5101
« on: March 16, 2018, 08:19:34 PM »
a. Express the general solution of the given system of equations in terms of real-value functions.
b. Also draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as $t\to \infty$.
$$\mathbf{x}' =\begin{pmatrix}
-1 &-4\\
1 &-1
\end{pmatrix}\mathbf{x}$$

Cheng Sheng

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Re: Q6--T5101
« Reply #1 on: March 17, 2018, 12:57:53 AM »
The steps are shown below.

Victor Ivrii

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Re: Q6--T5101
« Reply #2 on: March 17, 2018, 05:00:30 AM »
You found a bit more than required: not only the general solution but also fundamental matrix.

On the other hand, where is Part b?

Junya Zhang

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Re: Q6--T5101
« Reply #3 on: March 18, 2018, 11:56:12 AM »
Cheng Sheng's solution is correct, but here's the typed solution :)
a)
Let $$P=\begin{pmatrix} -1 & -4 \\ 1 & -1\end{pmatrix}$$
Characteristic polynomial:
$$\chi_P(\lambda) = \det\begin{pmatrix} -1-\lambda & -4 \\ 1 & -1-\lambda\end{pmatrix}=(\lambda +1)^2 +4$$
Thus, $$\lambda_1 = -1+2i, \lambda_2 = -1-2i$$
Consider $\lambda_1 =-1+2i$.
$$N(P- (-1+2i)I) = N\begin{pmatrix} -1+1-2i & -4 \\ 1 & -1+1-2i\end{pmatrix} = N\begin{pmatrix} -2i & -4 \\ 1 & -2i\end{pmatrix}= span\{\begin{pmatrix} 2i \\ 1\end{pmatrix}\}$$
Consider $$e^{(-1+2i)t}\begin{pmatrix} 2i \\ 1\end{pmatrix} =e^{-t} (\cos(2t) + i\sin(2t))\begin{pmatrix} 2i \\ 1\end{pmatrix}\ = e^{-t}\begin{pmatrix} 2i\cos(2t) -2\sin(2t) \\ \cos(2t) + i\sin(2t) \end{pmatrix} = e^{-t}\begin{pmatrix} -2\sin(2t) \\ \cos(2t) \end{pmatrix} + i  e^{-t}\begin{pmatrix} 2\cos(2t) \\ \sin(2t) \end{pmatrix}$$
Thus, the general solution of this system is
$$\textbf{x}(t) = c_1 e^{-t}\begin{pmatrix} -2\sin(2t) \\ \cos(2t) \end{pmatrix} + c_2  e^{-t}\begin{pmatrix} 2\cos(2t) \\ \sin(2t) \end{pmatrix}$$



b)
As $t\to\infty$, solution asymptotically converges to the origin in counter clockwise direction in spiral shapes.
See attached image.
« Last Edit: March 18, 2018, 12:01:14 PM by Junya Zhang »