# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Term Test 2 => Topic started by: Victor Ivrii on November 20, 2018, 05:46:11 AM

Title: TT2-P3
Post by: Victor Ivrii on November 20, 2018, 05:46:11 AM
(a) Find the general solution of
$$\mathbf{x}'=\begin{pmatrix}\hphantom{-}2 &\hphantom{-}1\\ -3 &-2\end{pmatrix}\mathbf{x}.$$

(b) Sketch corresponding trajectories. Describe the picture (stable/unstable, node, focus, center, saddle).

(c) Solve
$$\mathbf{x}'=\begin{pmatrix}\hphantom{-}2 &\hphantom{-}1\\ -3 &-2\end{pmatrix}\mathbf{x} + \begin{pmatrix}\hphantom{-} \frac{4}{e^t+e^{-t}} \\ -\frac{12}{e^t+e^{-t}}\end{pmatrix},\qquad \mathbf{x}(0)=\begin{pmatrix} 0 \\ 0\end{pmatrix}.$$
Title: Re: TT2-P3
Post by: Boyu Zheng on November 20, 2018, 07:39:51 AM
Title: Re: TT2-P3
Post by: Zhanhao Ye on November 20, 2018, 09:51:51 AM
The attachment is my solution.
Title: Re: TT2-P3
Post by: Jingze Wang on November 20, 2018, 10:05:33 AM
a)First, try to find the eigenvalues with respect to the parameter

$A=\begin{bmatrix} 2&1\\ -3&-2\\ \end{bmatrix}$

$det(A-rI)=(2-r)/(-2-r)+3=0$

$r^2-1=0$

$r=\pm1$

When r=-1, $3x_1+x_2=0$
We get $\begin{pmatrix}\hphantom{-} {1 }\\{-3}\end{pmatrix}$ is the corresponding eigenvector
When r=1,  $x_1+x_2=0$
We get $\begin{pmatrix}\hphantom{-} {1 }\\{-1}\end{pmatrix}$ is the corresponding eigenvector
Then the general solution is $$y=c_1\begin{pmatrix}\hphantom{-} {1 }\\{-3}\end{pmatrix}e^{-t}+c_2\begin{pmatrix}\hphantom{-} {1 }\\{-1}\end{pmatrix}e^{t}$$
Title: Re: TT2-P3
Post by: Michael Poon on November 20, 2018, 12:12:59 PM
Seems like no one has added a phase portrait yet.

I attached it below, it is a saddle, with eigenvalues real and opposite.
Title: Re: TT2-P3
Post by: Jingze Wang on November 20, 2018, 02:35:45 PM
Hi, my answer is same like yours except I got $C_1=-2ln2, C_2=0$
I am not confident about my answer, feel free to correct me :)
Title: Re: TT2-P3
Post by: Tzu-Ching Yen on November 20, 2018, 02:47:25 PM
I agree with Jingze's solution. It is obvious that Boyu's answer is not (0, 0) at $t =0$
Title: Re: TT2-P3
Post by: Yulin WANG on November 20, 2018, 04:03:36 PM
Hi, my answer is same like yours except I got $C_1=-2ln2, C_2=0$
I am not confident about my answer, feel free to correct me :)
My answer is the same as yours.
Title: Re: TT2-P3
Post by: Zhihong Yin on November 20, 2018, 04:11:24 PM
I am not sure.
Title: Re: TT2-P3
Post by: Siran Wang on November 22, 2018, 02:24:55 PM
a)
\begin{equation*}
A-\lambda I=\begin{pmatrix}
2-\lambda & 1\\
-3 & -2-\lambda
\end{pmatrix}
\end{equation*}

\begin{equation*}
\det(A-\lambda I)=(2-\lambda)(-2-\lambda)+3=0
\end{equation*}
\begin{equation*}
\end{equation*}
\begin{equation*}
\lambda_1=1~~~~\lambda_2=-1
\end{equation*}

When $\lambda_1=1$
\begin{equation*}
x_1+x_2=0
\end{equation*}
The first eigenvector is $\begin{pmatrix} 1\\ -1 \end{pmatrix}$

When $\lambda_1=-1$
\begin{equation*}
3x_1+x_2=0
\end{equation*}
The second eigenvector is $\begin{pmatrix} 1\\ -3 \end{pmatrix}$
So, the general solution is $y=C_1e^t\begin{pmatrix} 1\\ -1 \end{pmatrix}+C_2e^{-t}\begin{pmatrix} 1\\ -3 \end{pmatrix}$

graph in attachment

c)
\begin{equation*}
\phi=\begin{pmatrix}
e^t&e{-t}\\
-e^t&-3e^{-t}
\end{pmatrix}
\end{equation*}
\begin{equation*}
\begin{pmatrix}
e^t&e{-t}\\
-e^t&-3e^{-t}
\end{pmatrix}\begin{pmatrix}
U_1\\
U_2
\end{pmatrix}=\begin{pmatrix}
\frac{4}{e^t+e^{-t}}\\
\frac{-12}{e^t+e^{-t}}
\end{pmatrix}
\end{equation*}
\begin{equation*}
e^tU_1+e^{-t}U_2=\frac{4}{e^t+e^{-t}}
\end{equation*}
\begin{equation*}
-e^tU_1-3e^{-t}U_2= \frac{-12}{e^t+e^{-t}}
\end{equation*}
\begin{equation*}
U_2= \frac{4}{1+e^{-2t}}-e^{2t}U_1
\end{equation*}
\begin{equation*}
-e^tU_1-3e^{-t}(\frac{4}{1+e^{-2t}}-e^{2t}U_1)=\frac{-12}{e^t+e^{-t}}
\end{equation*}
\begin{equation*}
U_1=0~~~U_2=\frac{4}{1+e^{-2t}}
\end{equation*}
\begin{equation*}
V_1=\int U_1dt=\int0dt=0
\end{equation*}
\begin{equation*}
V_2=\int U_2dt=\int\frac{4}{1+e^{-2t}}dt=2ln(1+e^{2t})
\end{equation*}
\begin{equation*}\begin{pmatrix}
x_1\\
x_2
\end{pmatrix}=\begin{pmatrix}
C_1e^t+C_2e^{-t}+2e{-t}ln(1+e^{2t}\\
-C_1e^t-3C_2e^{-t}-6e{-t}ln(1+e^{2t}
\end{pmatrix}
\end{equation*}
since $\begin{equation*}x(0)=\begin{pmatrix} 0\\ 0 \end{pmatrix} \end{equation*}$
\begin{equation*}
C_1+C_2+2ln2=0~~~-C_1-3C_2-6ln2=0
\end{equation*}\begin{equation*}
C_1=0~~~C_2=-2ln2
\end{equation*}
Title: Re: TT2-P3
Post by: Victor Ivrii on November 25, 2018, 11:14:58 AM
There is computer-generated picture