# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Term Test 2 => Topic started by: Victor Ivrii on November 19, 2019, 04:24:35 AM

Title: Problem 4 (noon)
Post by: Victor Ivrii on November 19, 2019, 04:24:35 AM
Find the general real solution to
$$\mathbf{x}'=\begin{pmatrix} 1 & 3\\ -2 &-3\end{pmatrix}\mathbf{x}$$
and sketch trajectories.
Title: Re: Problem 4 (noon)
Post by: Yuying Chen on November 19, 2019, 05:17:25 AM
$\det(A-\lambda I)=0\\$
$\begin{vmatrix} 1-\lambda & 3 \\ -2 & -3-\lambda\\ \end{vmatrix}=(1-\lambda)(-3-\lambda)+6=0\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\quad \therefore \lambda=-1\pm \sqrt 2 i\\$
$\text{when$\lambda =-1+ \sqrt 2 i$},\\$
$\begin{pmatrix} 2-\sqrt 2 i & 3 \\ -2 & -2-\sqrt 2 i \end{pmatrix}= \begin{pmatrix} 2+\sqrt 2 i \\ -2 \end{pmatrix}\\$
$e^{(-1+ \sqrt 2 i)t}\begin{pmatrix} 2+\sqrt 2 i \\ -2 \end{pmatrix}=e^{-t}\begin{pmatrix} 2+\sqrt 2 i \\ -2 \end{pmatrix}(\cos\sqrt2t+i\sin\sqrt 2t)\\$
$\qquad\qquad\qquad\qquad =e^{-t}\begin{pmatrix} 2\cos\sqrt 2t+2i\sin\sqrt2t+\sqrt2i\cos\sqrt2t-\sqrt2\sin\sqrt2t \\ -2\cos\sqrt2t-2i\sin\sqrt2t \end{pmatrix}\\$
$\qquad\qquad\qquad\qquad =e^{-t}i\begin{pmatrix} 2\sin\sqrt2t+\sqrt2\cos\sqrt2t\\ -2\sin\sqrt2t \end{pmatrix}+e^{-t}\begin{pmatrix} 2\cos\sqrt 2t -\sqrt2\sin\sqrt2t\\ -2\cos\sqrt2t \end{pmatrix}\\$
$\therefore x(t)=c_1e^{-t}\begin{pmatrix} 2\sin\sqrt2t+\sqrt2\cos\sqrt2t\\ -2\sin\sqrt2t \end{pmatrix}+c_2e^{-t}\begin{pmatrix} 2\cos\sqrt 2t -\sqrt2\sin\sqrt2t \\ -2\cos\sqrt2t \end{pmatrix}\\$
Title: Re: Problem 4 (noon)
Post by: Ziqian Qiu on November 19, 2019, 05:24:11 AM
this is my solution
Title: Re: Problem 4 (noon)
Post by: Ziqian Qiu on November 19, 2019, 05:30:32 AM
nvm I just updated the previous post
Title: Re: Problem 4 (noon)
Post by: xuanzhong on November 19, 2019, 05:47:01 AM
here's the solution including sketching.
Title: Re: Problem 4 (noon)
Post by: Changhao Jiang on November 19, 2019, 05:50:48 AM
To find eigenvalues, $det(A-\lambda x)=(1-\lambda)(-3-\lambda)-(-2)(3)=0$,we can get $\lambda = -1-\sqrt{2}$ or $\lambda = -1+\sqrt{2}$
To find eigenvectors, when $\lambda=-1-\sqrt{2}$,
$\begin{pmatrix} 2+\sqrt{2} & 3 \\ -2 & -2+\sqrt{2} \end{pmatrix} ~ \begin{pmatrix} 2 & 2-\sqrt{2} \\ 0 & 0 \end{pmatrix}$
the eigenvector is $\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix}$
so $e^{(-1-\sqrt{2})t}\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix} = e^{-t}(\cos\sqrt{2}t-i\sin\sqrt{2}t)\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix} = e^{-t} (\begin{bmatrix} 2\cos\sqrt{2}t-\sqrt{2}cos\sqrt{2}t \\ -2cos\sqrt{2}t \end{bmatrix} + i\begin{bmatrix} -2\sin\sqrt{2}t+\sqrt{2}sin\sqrt{2}t \\ 2sin\sqrt{2}t \end{bmatrix})$
therefore, the general solution is $x(t)=c_1 e^{-t} \begin{bmatrix} 2\cos\sqrt{2}t-\sqrt{2}cos\sqrt{2}t \\ -2cos\sqrt{2}t \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} -2\sin\sqrt{2}t+\sqrt{2}sin\sqrt{2}t \\ 2sin\sqrt{2}t \end{bmatrix}$
Title: Re: Problem 4 (noon)
Post by: NANAC on November 19, 2019, 09:21:22 AM
Title: Re: Problem 4 (noon)
Post by: baixiaox on November 19, 2019, 05:46:00 PM
Title: Re: Problem 4 (noon)
Post by: Mingdi Xie on November 20, 2019, 03:07:38 PM
This is my solution. :)
Title: Re: Problem 4 (noon)
Post by: Victor Ivrii on November 24, 2019, 11:04:24 AM
What everybody is missing

we see that characteristic roots $k_{1,2}=-1\pm \sqrt{2}i$ are complex, with negative real part. So, it is  stable focus  and with  clock-wise  orientation  since the bottom-left element is negative.