MAT244--2019F > Term Test 2

Problem 4 (noon)

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Victor Ivrii:
Find the general real solution to
$$\mathbf{x}'=\begin{pmatrix} 1 & 3\\ -2 &-3\end{pmatrix}\mathbf{x}$$
and sketch trajectories.

Yuying Chen:
$\det(A-\lambda I)=0\\$
$\begin{vmatrix} 1-\lambda & 3 \\ -2 & -3-\lambda\\ \end{vmatrix}=(1-\lambda)(-3-\lambda)+6=0\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\quad \therefore \lambda=-1\pm \sqrt 2 i\\$
$\text{when$\lambda =-1+ \sqrt 2 i$},\\$
$\begin{pmatrix} 2-\sqrt 2 i & 3 \\ -2 & -2-\sqrt 2 i \end{pmatrix}= \begin{pmatrix} 2+\sqrt 2 i \\ -2 \end{pmatrix}\\$
$e^{(-1+ \sqrt 2 i)t}\begin{pmatrix} 2+\sqrt 2 i \\ -2 \end{pmatrix}=e^{-t}\begin{pmatrix} 2+\sqrt 2 i \\ -2 \end{pmatrix}(\cos\sqrt2t+i\sin\sqrt 2t)\\$
$\qquad\qquad\qquad\qquad =e^{-t}\begin{pmatrix} 2\cos\sqrt 2t+2i\sin\sqrt2t+\sqrt2i\cos\sqrt2t-\sqrt2\sin\sqrt2t \\ -2\cos\sqrt2t-2i\sin\sqrt2t \end{pmatrix}\\$
$\qquad\qquad\qquad\qquad =e^{-t}i\begin{pmatrix} 2\sin\sqrt2t+\sqrt2\cos\sqrt2t\\ -2\sin\sqrt2t \end{pmatrix}+e^{-t}\begin{pmatrix} 2\cos\sqrt 2t -\sqrt2\sin\sqrt2t\\ -2\cos\sqrt2t \end{pmatrix}\\$
$\therefore x(t)=c_1e^{-t}\begin{pmatrix} 2\sin\sqrt2t+\sqrt2\cos\sqrt2t\\ -2\sin\sqrt2t \end{pmatrix}+c_2e^{-t}\begin{pmatrix} 2\cos\sqrt 2t -\sqrt2\sin\sqrt2t \\ -2\cos\sqrt2t \end{pmatrix}\\$

Ziqian Qiu:
this is my solution

Ziqian Qiu:
nvm I just updated the previous post

xuanzhong:
here's the solution including sketching.