Author Topic: Q6--T0901  (Read 1673 times)

Victor Ivrii

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Q6--T0901
« on: March 16, 2018, 08:17:56 PM »
a. Express the general solution of the given system of equations in terms of real-valued functions.
b. Also draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as $t\to \infty$.
$$\mathbf{x}' =\begin{pmatrix}
4 &-3\\
8 &-6
\end{pmatrix}\mathbf{x}$$

Vivian Ngo

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Re: Q6--T0901
« Reply #1 on: March 18, 2018, 01:30:38 AM »
Solving for eigenvalues:
\begin{gather*}
(4-r)(-6-r)+24 = 0\\
\implies - 24 - 4r + 6r + r^2 = 0\\
\implies r^2 + 2r = 0 \implies r(r+2) = 0
\end{gather*}
eigenvalues are $0$ and $-2$.

When $r = 0$,  Nullspace of $\begin{pmatrix}
    4 & -3 \\
    8 & -6
  \end{pmatrix}$ equals to  Nullspace of $\begin{pmatrix}
    4 & -3 \\
    0 & 0
  \end{pmatrix}$, which is equal to the span of  $\begin{pmatrix}
    3 \\
    4
  \end{pmatrix}$.
When $r = -2$,  Nullspace of $\begin{pmatrix}
    6 & -3 \\
    8 & -4
  \end{pmatrix}$, which is equal to Nullspace of $\begin{pmatrix}
    6 & -3 \\
    0 & 0
  \end{pmatrix}$, which is a span of $\begin{pmatrix}
    1 \\
    2
  \end{pmatrix}$.

Then
$$\mathbf{x} = c_1\begin{pmatrix}3\\4\end{pmatrix} + c_2e^{-2t}\begin{pmatrix}1\\2\end{pmatrix}$$

As $t$ goes to infinity, the solutions tend to $c_1\begin{pmatrix}3\\4\end{pmatrix}$. A sketch of the phase portrait:


« Last Edit: March 18, 2018, 04:25:47 AM by Victor Ivrii »

Victor Ivrii

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Re: Q6--T0901
« Reply #2 on: March 18, 2018, 04:14:25 AM »
Vivian, please learn LaTeX. I fixed it foe you

And do not use external web servers for images. Use attachments.
« Last Edit: March 18, 2018, 04:26:50 AM by Victor Ivrii »