MAT244--2019F > Term Test 2

Problem 4 (main sitting)

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Ruodan Chen:
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

yueyangyu:
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}) $

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

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

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

zhaorola:

--- Quote from: Aparna on November 19, 2019, 08:47:30 AM ---Computer-generated sketch:

--- End quote ---

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

nayan:

--- Quote from: zhaorola on November 19, 2019, 01:25:45 PM ---
--- Quote from: Aparna on November 19, 2019, 08:47:30 AM ---Computer-generated sketch:

--- End quote ---

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

--- End quote ---

Yes this appears to be the correct phase portrait.

Mingdi Xie:
Here is my solution

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