Toronto Math Forum
MAT244--2018F => MAT244--Tests => Term Test 2 => Topic started by: Victor Ivrii on November 20, 2018, 05:46:11 AM
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(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
$$
\mathbf{x}'=\begin{pmatrix}\hphantom{-}2 &\hphantom{-}1\\
-3 &-2\end{pmatrix}\mathbf{x} +
\begin{pmatrix}\hphantom{-} \frac{4}{e^t+e^{-t}} \\
-\frac{12}{e^t+e^{-t}}\end{pmatrix},\qquad
\mathbf{x}(0)=\begin{pmatrix} 0 \\
0\end{pmatrix}.
$$
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here is my answer
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The attachment is my solution.
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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 $$y=c_1\begin{pmatrix}\hphantom{-} {1 }\\{-3}\end{pmatrix}e^{-t}+c_2\begin{pmatrix}\hphantom{-} {1 }\\{-1}\end{pmatrix}e^{t}$$
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Seems like no one has added a phase portrait yet.
I attached it below, it is a saddle, with eigenvalues real and opposite.
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here is my answer
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 :)
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I agree with Jingze's solution. It is obvious that Boyu's answer is not (0, 0) at $t =0$
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here is my answer
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.
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I am not sure.
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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 $y=C_1e^t\begin{pmatrix}
1\\
-1
\end{pmatrix}+C_2e^{-t}\begin{pmatrix}
1\\
-3
\end{pmatrix}$
b) unstable, saddle point
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*}
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There is computer-generated picture