# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Term Test 2 => Topic started by: Victor Ivrii on November 19, 2019, 04:23:20 AM

Title: Problem 4 (main sitting)
Post by: Victor Ivrii on November 19, 2019, 04:23:20 AM
Find the general real solution to
$$\mathbf{x}'=\begin{pmatrix} 3 & 3\\ -2 &-1\end{pmatrix}\mathbf{x}$$
and sketch trajectories.
Title: Re: Problem 4 (main sitting)
Post by: Yiheng Bian on November 19, 2019, 04:30:28 AM
$$det(A-{\lambda}I)=0\\ \begin{vmatrix} 3-\lambda & 3 \\ -2 & -1-\lambda \end{vmatrix}={\lambda -1}^2+2=0$$
WRONG So
$$\lambda_1=1 +\sqrt{2}i\\ \lambda_2=1-\sqrt{2}i$$
$$\text{when } \lambda=1 +\sqrt{2}i\\ \begin{vmatrix} 2-\sqrt{2}i & 3 \\ -2 & -2-\sqrt{2}i \end{vmatrix} = \begin{vmatrix} 0 \\ 0 \end{vmatrix}$$
RREF:
$$\begin{pmatrix} 2-\sqrt{2}i & 3 \\ 0 & 0 \end{pmatrix} \quad = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad$$

Let x_2=t
So we can get:
$${(2-\sqrt{2}i})x_1=-3x_2=-3t\\ x_1=\frac{-3t}{2-\sqrt{2}i}$$
So
$$t*\begin{pmatrix} -1-\frac{\sqrt{2}i}{2} \\ 1 \end{pmatrix} \quad$$
Therefore:
$$e^{1+i\sqrt2t}\begin{pmatrix} -1-\frac{\sqrt{2}i}{2} \\ 1 \end{pmatrix} \quad = e^t[\begin{pmatrix} -cos(\sqrt{2}t)+\frac{\sqrt{2}}{2}*sin(\sqrt{2}t) \\ cos(\sqrt{2}t) \end{pmatrix} \quad + i\begin{pmatrix} -sin(\sqrt{2}t)-\frac{\sqrt{2}}{2}*cos(\sqrt{2}t) \\ sin(\sqrt{2}t) \end{pmatrix} \quad]$$
So, general solution:
$$y=c_1e^t\begin{pmatrix} -cos(\sqrt{2}t)+\frac{\sqrt{2}}{2}*sin(\sqrt{2}t) \\ cos(\sqrt{2}t) \end{pmatrix} \quad + c_2e^t\begin{pmatrix} -sin(\sqrt{2}t)-\frac{\sqrt{2}}{2}*cos(\sqrt{2}t) \\ sin(\sqrt{2}t) \end{pmatrix} \quad$$
Title: Re: Problem 4 (main sitting)
Post by: Lisa Dou on November 19, 2019, 05:17:54 AM
The Sketch
Title: Re: Problem 4 (main sitting)
Post by: Aparna on November 19, 2019, 08:47:30 AM
Computer-generated sketch:
Title: Re: Problem 4 (main sitting)
Post by: Sifan Shao on November 19, 2019, 09:15:01 AM
The Sketch
we can do root2i*R1+R2 to simplify the matrix
Title: Re: Problem 4 (main sitting)
Post by: Ruodan Chen on November 19, 2019, 10:41:19 AM
4) $x'=\left(\begin{array}{cc} 3 & 3\\ -2 & -1 \end{array}\right)x$

a)

$det(A-\lambda I)=det(\begin{array}{cc} 3-\lambda & 3\\ -2 & -1-\lambda \end{array})=(3-\lambda)(-1-\lambda)+6=0$

$\lambda^{2}-2\lambda+3=0$

$(\lambda-1)^{2}=-2$

\$\lambda=1\pm\sqrt{2}i$

\$\lambda=1+\sqrt{2}i$

$(\begin{array}{ccc} 2-\sqrt{2}i & 3 & 0\\ -2 & -2-\sqrt{2}i & 0 \end{array}) = (\begin{array}{ccc} 2-\sqrt{2}i & 3 & 0\\ 0 & 0 & 0 \end{array})$

$v=(\begin{array}{c}2+\sqrt{2}i\\-2\end{array})$

$e^{(1+\sqrt{2}i)t}(\begin{array}{c} 2+\sqrt{2}i\\ -2 \end{array})=e^{t}(cos\sqrt{2}t+isin\sqrt{2}t)(\begin{array}{c}2+\sqrt{2}i\\-2\end{array}) = e^{t}(\begin{array}{c} 2cos\sqrt{2}t+2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\ -2cos\sqrt{2}t-2isin\sqrt{2}t \end{array})=e^{t}(\begin{array}{c} 2cos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\ -2cos\sqrt{2}t \end{array})+ie^{t}(\begin{array}{c} 2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t\\ -2isin\sqrt{2}t \end{array})$

$x(t)=c_{1}e{}^{t}(\begin{array}{c} 2cos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\ -2cos\sqrt{2}t \end{array})+c_{2}e{}^{t}(\begin{array}{c} 2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t\\ -2isin\sqrt{2}t \end{array})$
You should consider real solutions
Title: Re: Problem 4 (main sitting)
Post by: yueyangyu on November 19, 2019, 12:52:21 PM
4) $x'=\left(\begin{array}{cc} 3 & 3\\ -2 & -1 \end{array}\right)x$

a)

$det(A-\lambda I)=det(\begin{array}{cc} 3-\lambda & 3\\ -2 & -1-\lambda \end{array})=(3-\lambda)(-1-\lambda)+6=0$

$\lambda^{2}-2\lambda+3=0$

$(\lambda-1)^{2}=-2$

$\lambda=1\pm\sqrt{2}i$

$\lambda=1+\sqrt{2}i$

$(\begin{array}{ccc} 2-\sqrt{2}i & 3 & 0\\ -2 & -2-\sqrt{2}i & 0 \end{array}) = (\begin{array}{ccc} 2-\sqrt{2}i & 3 & 0\\ 0 & 0 & 0 \end{array})$

$v=(\begin{array}{c}2+\sqrt{2}i\\-2\end{array})$

$e^{(1+\sqrt{2}i)t}(\begin{array}{c} 2+\sqrt{2}i\\ -2 \end{array})=e^{t}(cos\sqrt{2}t+isin\sqrt{2}t)(\begin{array}{c}2+\sqrt{2}i\\-2\end{array}) = e^{t}(\begin{array}{c} 2cos\sqrt{2}t+2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\ -2cos\sqrt{2}t-2isin\sqrt{2}t \end{array})=e^{t}(\begin{array}{c} 2cos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\ -2cos\sqrt{2}t \end{array})+ie^{t}(\begin{array}{c} 2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t\\ -2isin\sqrt{2}t \end{array})$

$x(t)=c_{1}e{}^{t}(\begin{array}{c} 2cos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\ -2cos\sqrt{2}t \end{array})+c_{2}e{}^{t}(\begin{array}{c} 2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t\\ -2isin\sqrt{2}t \end{array})$
Title: Re: Problem 4 (main sitting)
Post by: zhaorola on November 19, 2019, 01:25:45 PM
Computer-generated sketch:

This doesn't look right. It should instead be an unstable spiral. I've attached an accurate phase portrait.
Title: Re: Problem 4 (main sitting)
Post by: nayan on November 19, 2019, 01:53:29 PM
Computer-generated sketch:

This doesn't look right. It should instead be an unstable spiral. I've attached an accurate phase portrait.

Yes this appears to be the correct phase portrait.
Title: Re: Problem 4 (main sitting)
Post by: Mingdi Xie on November 19, 2019, 09:58:57 PM
Here is my solution
Title: Re: Problem 4 (main sitting)
Post by: Victor Ivrii on November 24, 2019, 10:45:37 AM
What everybody is missing

it is  unstable focus  and with  clock-wise  orientation  since the bottom-left element is negative.