Author Topic: TT2B-P4  (Read 990 times)

Victor Ivrii

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TT2B-P4
« on: November 20, 2018, 05:56:49 AM »
(a) Find the general real solution to
$$
\mathbf{x}'=\begin{pmatrix}
-3 &-2\\
\hphantom{-}5 &-5\end{pmatrix}\mathbf{x}.$$
(b)  Sketch trajectories. Describe the picture (stable/unstable, node, focus, center, saddle).

Xiaoyuan Wang

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Re: TT2B-P4
« Reply #1 on: November 20, 2018, 06:04:04 AM »
Answer

Yulin WANG

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Re: TT2B-P4
« Reply #2 on: November 20, 2018, 03:10:39 PM »
(a)
\begin{align*}
Let A &= \begin{bmatrix}
-3 & -2 \\
5 & -5
\end{bmatrix}\\
~\\
A - \lambda I &= \begin{bmatrix}
-3 - \lambda & -2 \\
5 & -5 - \lambda
\end{bmatrix}\\
~\\
det(A - \lambda I) &= (5 + \lambda)(3 + \lambda) + 10\\
~\\
&= \lambda^{2} + 8\lambda +25\\
~\\
&= (\lambda +4)^{2} + 9 = 0\\
~\\
\lambda &= -4 \pm 3i \\
~\\
For \ \lambda = -4 + 3i, A - \lambda I &= \begin{bmatrix}
1 - 3i & -2 \\
5 & -1 - 3i
\end{bmatrix}\\
~\\
Since,\ null(\begin{bmatrix}
1 - 3i & -2 \\
5 & -1 - 3i
\end{bmatrix}) &= span\{\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\}
~\\
So \ the \ eigenvector\  v &= \begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
Since \ e^{\lambda t}v &= e^{-4 + 3i}\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
&= e^{-4t}e^{3it}\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
&= e^{-4t}(cos3t + isin3t)\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
&= e^{-4t}\begin{bmatrix}
3icos3t - 3sin3t + cos3t + isin3t\\
5cos3t +5isin3t \\
\end{bmatrix}\\
~\\
So, \ \phi_{1}(t) &= e^{-4t}\begin{bmatrix}
- 3sin3t + cos3t\\
5cos3t\\
\end{bmatrix}\\
~\\
\phi_{2}(t) &= e^{-4t}\begin{bmatrix}
3cos3t + sin3t\\
5sin3t \\
\end{bmatrix}\\
~\\
Thus, \ x(t) &= c_{1}\phi_{1}(t) + c_{2}\phi_{2}(t)\\
~\\
&= c_{1}e^{-4t}\begin{bmatrix}
- 3sin3t + cos3t\\
5cos3t\\
\end{bmatrix} + c_{2}e^{-4t}\begin{bmatrix}
3cos3t + sin3t\\
5sin3t \\
\end{bmatrix}\\
\end{align*}
(b) In the attachment.
« Last Edit: November 20, 2018, 08:05:07 PM by Yulin Wang »

Jingze Wang

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Re: TT2B-P4
« Reply #3 on: November 20, 2018, 03:37:20 PM »
\begin{align*}
Let A &= \begin{bmatrix}
-3 & -2 \\
5 & -5
\end{bmatrix}\\
~\\
A - \lambda I &= \begin{bmatrix}
-3 - \lambda & -2 \\
5 & -5 - \lambda
\end{bmatrix}\\
~\\
det(A - \lambda I) &= (5 + \lambda)(3 + \lambda) + 10\\
~\\
&= \lambda^{2} + 8\lambda +25\\
~\\
&= (\lambda +4)^{2} + 9 = 0\\
~\\
\lambda &= -4 \pm 3i \\
~\\
For \lambda = -4 + 3i, A - \lambda I &= \begin{bmatrix}
1 - 3i & -2 \\
5 & 1 - 3i
\end{bmatrix}\\
~\\
Since,\ nul(\begin{bmatrix}
1 - 3i & -2 \\
5 & 1 - 3i
\end{bmatrix}) &= span\{\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\}
~\\
So \ the \ eigenvector\  v &= \begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
Since \ e^{\lambda t}v &= e^{-4 + 3i}\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
&= e^{-4t}e^{3it}\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
&= e^{-4t}(cos3t + isin3t)\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
&= e^{-4t}\begin{bmatrix}
3icos3t - 3sin3t + cos3t + isin3t\\
5cos3t +5isin3t \\
\end{bmatrix}\\
~\\
So, \ \phi_{1}(t) &= e^{-4t}\begin{bmatrix}
- 3sin3t + cos3t\\
5cos3t\\
\end{bmatrix}\\
~\\
\phi_{2}(t) &= e^{-4t}\begin{bmatrix}
3cos3t + sin3t\\
5sin3t \\
\end{bmatrix}\\
~\\
Thus, \ x(t) &= c_{1}\phi_{1}(t) + c_{2}\phi_{2}(t)\\
~\\
&= c_{1}e^{-4t}\begin{bmatrix}
- 3sin3t + cos3t\\
5cos3t\\
\end{bmatrix} + c_{2}e^{-4t}\begin{bmatrix}
3cos3t + sin3t\\
5sin3t \\
\end{bmatrix}\\
\end{align*}
(b) In the attachment.
Hi Yulin, I think you made a mistake in your graph, since c=5>0, your graph should be counterclockwise instead of clockwise

Yulin WANG

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Re: TT2B-P4
« Reply #4 on: November 20, 2018, 03:50:55 PM »
\begin{align*}
Let A &= \begin{bmatrix}
-3 & -2 \\
5 & -5
\end{bmatrix}\\
~\\
A - \lambda I &= \begin{bmatrix}
-3 - \lambda & -2 \\
5 & -5 - \lambda
\end{bmatrix}\\
~\\
det(A - \lambda I) &= (5 + \lambda)(3 + \lambda) + 10\\
~\\
&= \lambda^{2} + 8\lambda +25\\
~\\
&= (\lambda +4)^{2} + 9 = 0\\
~\\
\lambda &= -4 \pm 3i \\
~\\
For \lambda = -4 + 3i, A - \lambda I &= \begin{bmatrix}
1 - 3i & -2 \\
5 & 1 - 3i
\end{bmatrix}\\
~\\
Since,\ nul(\begin{bmatrix}
1 - 3i & -2 \\
5 & 1 - 3i
\end{bmatrix}) &= span\{\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\}
~\\
So \ the \ eigenvector\  v &= \begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
Since \ e^{\lambda t}v &= e^{-4 + 3i}\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
&= e^{-4t}e^{3it}\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
&= e^{-4t}(cos3t + isin3t)\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
&= e^{-4t}\begin{bmatrix}
3icos3t - 3sin3t + cos3t + isin3t\\
5cos3t +5isin3t \\
\end{bmatrix}\\
~\\
So, \ \phi_{1}(t) &= e^{-4t}\begin{bmatrix}
- 3sin3t + cos3t\\
5cos3t\\
\end{bmatrix}\\
~\\
\phi_{2}(t) &= e^{-4t}\begin{bmatrix}
3cos3t + sin3t\\
5sin3t \\
\end{bmatrix}\\
~\\
Thus, \ x(t) &= c_{1}\phi_{1}(t) + c_{2}\phi_{2}(t)\\
~\\
&= c_{1}e^{-4t}\begin{bmatrix}
- 3sin3t + cos3t\\
5cos3t\\
\end{bmatrix} + c_{2}e^{-4t}\begin{bmatrix}
3cos3t + sin3t\\
5sin3t \\
\end{bmatrix}\\
\end{align*}
(b) In the attachment.
Hi Yulin, I think you made a mistake in your graph, since c=5>0, your graph should be counterclockwise instead of clockwise
Thanks for your help.

Victor Ivrii

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Re: TT2B-P4
« Reply #5 on: November 25, 2018, 11:44:00 AM »
Computer generated