# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Quiz-6 => Topic started by: Victor Ivrii on March 16, 2018, 08:16:03 PM

Title: Q6--T0701
Post by: Victor Ivrii on March 16, 2018, 08:16:03 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 $t\to \infty$.
$$\mathbf{x}' =\begin{pmatrix} 1 &2\\ -5 &-1 \end{pmatrix}\mathbf{x}$$
Title: Re: Q6--T0701
Post by: Cheng Sheng on March 17, 2018, 01:58:40 AM
The steps are shown below.
Title: Re: Q6--T0701
Post by: Victor Ivrii on March 17, 2018, 05:13:38 AM
Why this colour of your attachment? Make it black and white.

And where is a plot?
Title: Re: Q6--T0701
Post by: Junya Zhang on March 18, 2018, 11:20:29 AM
Cheng Sheng's solution is correct, but here's the typed solution :)
a)
Let $$P=\begin{pmatrix} 1 & 2 \\ -5 & -1\end{pmatrix}$$
Characteristic polynomial:
$$\chi_P(\lambda) = det\begin{pmatrix} 1-\lambda & 2 \\ -5 & -1-\lambda\end{pmatrix}=\lambda^2 + 9$$
Thus, $$\lambda_1 = 3i, \lambda_2 = -3i$$
Consider $\lambda_1 = 3i$.
$$N(P-3iI) = N\begin{pmatrix} 1-3i & 2 \\ -5 & -1-3i\end{pmatrix} = span\{\begin{pmatrix} -2 \\ 1-3i\end{pmatrix}\}$$
Consider $$e^{3it}\begin{pmatrix} -2 \\ 1-3i\end{pmatrix} = \cos(3t) + i\sin(3t)\begin{pmatrix} -2 \\ 1-3i\end{pmatrix}\ = \begin{pmatrix} -2\cos(3t) -2i\sin(3t) \\ \cos(3t) + i\sin(3t) - 3i\cos(3t) + 3\sin(3t)\end{pmatrix} =\begin{pmatrix} -2\cos(3t) \\ \cos(3t) + 3\sin(3t)\end{pmatrix} + i \begin{pmatrix} -2\sin(3t) \\ \sin(3t) - 3\cos(3t)\end{pmatrix}$$
Thus, the general solution of this system is
$$\textbf{x}(t) = c_1\begin{pmatrix} -2\cos(3t) \\ \cos(3t) + 3\sin(3t)\end{pmatrix} + c_2 \begin{pmatrix} -2\sin(3t) \\ \sin(3t) - 3\cos(3t)\end{pmatrix}$$

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