MAT244--2018F > Term Test 2

TT2-P3

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Victor Ivrii:
(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
$$\mathbf{x}'=\begin{pmatrix}\hphantom{-}2 &\hphantom{-}1\\ -3 &-2\end{pmatrix}\mathbf{x} + \begin{pmatrix}\hphantom{-} \frac{4}{e^t+e^{-t}} \\ -\frac{12}{e^t+e^{-t}}\end{pmatrix},\qquad \mathbf{x}(0)=\begin{pmatrix} 0 \\ 0\end{pmatrix}.$$

Boyu Zheng:
here is my answer

Zhanhao Ye:
The attachment is my solution.

Jingze Wang:
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 $$y=c_1\begin{pmatrix}\hphantom{-} {1 }\\{-3}\end{pmatrix}e^{-t}+c_2\begin{pmatrix}\hphantom{-} {1 }\\{-1}\end{pmatrix}e^{t}$$

Michael Poon:
Seems like no one has added a phase portrait yet.

I attached it below, it is a saddle, with eigenvalues real and opposite.