Author Topic: TT2A-P3  (Read 924 times)

Victor Ivrii

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TT2A-P3
« on: November 20, 2018, 05:52:12 AM »
(a) Find the general solution of
$$
\mathbf{x}'=\begin{pmatrix} \ 4 & \ 1\\
-3 &0\end{pmatrix}\mathbf{x}.$$

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

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

Mallory Schneider

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Re: TT2A-P3
« Reply #1 on: November 20, 2018, 12:08:17 PM »
Part a) and b)

Michael Poon

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Re: TT2A-P3
« Reply #2 on: November 20, 2018, 12:17:52 PM »
I think you could also characterise the phase portrait as a node? (unstable node)

Mallory Schneider

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Re: TT2A-P3
« Reply #3 on: November 20, 2018, 12:27:36 PM »
Part C

Victor Ivrii

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Re: TT2A-P3
« Reply #4 on: November 25, 2018, 12:34:25 PM »
I think you could also characterise the phase portrait as a node? (unstable node)
Indeed


Computer generated