### Author Topic: TT2B-P4  (Read 5906 times)

#### Victor Ivrii

• Elder Member
• Posts: 2607
• Karma: 0
##### 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

• Jr. Member
• Posts: 8
• Karma: 9
##### Re: TT2B-P4
« Reply #1 on: November 20, 2018, 06:04:04 AM »

#### Yulin WANG

• Full Member
• Posts: 17
• Karma: 25
• MAT244H1 2018F
##### 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

• Full Member
• Posts: 30
• Karma: 25
##### 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.

#### Yulin WANG

• Full Member
• Posts: 17
• Karma: 25
• MAT244H1 2018F
##### 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.