Question

the solution of the linear system \\(\mathbf{x} = \mathbf{a}\mathbf{x}\\) is given.

\\(\mathbf{a} = \

$$\begin{pmatrix} -1 & -2 \\\\ 6 & 7 \\end{pmatrix}$$

\\), \\(\mathbf{x}(t) = c_1 \

$$\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}$$

e^t + c_2 \

$$\begin{pmatrix} -10 \\\\ 30 \\end{pmatrix}$$

e^{5t}\\)

please discuss the nature of the solution in a neighborhood of \\((0, 0)\\).

• if \\(\mathbf{x}(0) = \mathbf{x}_0\\) lies on the line \\(y = -x\\), then \\(\mathbf{x}(t)\\) approaches \\((0, 0)\\) along this line. otherwise \\(\mathbf{x}(t)\\) approaches \\((0, 0)\\) from the direction determined by \\(y = -3x\\).
• if \\(\mathbf{x}(0) = \mathbf{x}_0\\) lies on the line \\(y = -3x\\), then \\(\mathbf{x}(t)\\) approaches \\((0, 0)\\) along this line. otherwise \\(\mathbf{x}(t)\\) approaches \\((0, 0)\\) from the direction determined by \\(y = -x\\).
• if \\(\mathbf{x}(0) = \mathbf{x}_0\\) lies on the line \\(y = -x\\), then \\(\mathbf{x}(t)\\) becomes unbounded along this line. otherwise \\(\mathbf{x}(t)\\) becomes unbounded and \\(y = -3x\\) serves as an asymptote.
• if \\(\mathbf{x}(0) = \mathbf{x}_0\\) lies on the line \\(y = -3x\\), then \\(\mathbf{x}(t)\\) becomes unbounded along this line. otherwise \\(\mathbf{x}(t)\\) becomes unbounded and \\(y = -x\\) serves as an asymptote.
• all solutions spiral toward \\((0, 0)\\).

with the aid of a calculator or a cas, find and graph the solution that satisfies \\(\mathbf{x}(0) = (1, 1)\\). (enter any column vector as a row vector. select update graph to see your response plotted on the graph. select the submit button to grade your response.)

Answer:

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<reasoning>

Analyze eigenvalues and eigenvectors

Using the Linear Systems of ODEs knowledge point
\[
\mathbf{X}(t) = c_1

$$\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

e^t + c_2

$$\begin{pmatrix} -10 \\ 30 \end{pmatrix}$$

e^{5t}
\]
\[
\lambda_1 = 1 > 0, \quad \mathbf{v}_1 =

$$\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

\implies y = -x
\]
\[
\lambda_2 = 5 > 0, \quad \mathbf{v}_2 =

$$\begin{pmatrix} -10 \\ 30 \end{pmatrix}$$

\propto

$$\begin{pmatrix} 1 \\ -3 \end{pmatrix}$$

\implies y = -3x
\]

Determine asymptotic behavior as t increases

Using the Phase Portrait Analysis knowledge point
\[
\text{Since } \lambda_1 > 0 \text{ and } \lambda_2 > 0, \text{ both terms grow exponentially as } t \to \infty.
\]
\[
\text{If } c_2 = 0 \text{ (i.e., } \mathbf{X}(0) \text{ lies on } y = -x\text{), } \mathbf{X}(t) \text{ becomes unbounded along } y = -x.
\]
\[
\text{If } c_2
e 0, \text{ since } \lambda_2 = 5 > \lambda_1 = 1, \text{ the } e^{5t} \text{ term dominates as } t \to \infty.
\]
\[
\mathbf{X}(t) \approx c_2

$$\begin{pmatrix} -10 \\ 30 \end{pmatrix}$$

e^{5t} \implies y = -3x \text{ serves as an asymptote.}
\]

Solve the initial value problem

Using the Linear Systems of ODEs knowledge point
\[
\mathbf{X}(0) = c_1

$$\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

+ c_2

$$\begin{pmatrix} -10 \\ 30 \end{pmatrix}$$

=

$$\begin{pmatrix} 1 \\ 1 \end{pmatrix}$$

\]
\[

$$\begin{cases} c_1 - 10c_2 = 1 \\ -c_1 + 30c_2 = 1 \end{cases}$$

\]
\[
20c_2 = 2 \implies c_2 = \frac{1}{10}
\]
\[
c_1 = 1 + 10c_2 = 2
\]
\[
\mathbf{X}(t) = 2

$$\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

e^t + \frac{1}{10}

$$\begin{pmatrix} -10 \\ 30 \end{pmatrix}$$

e^{5t} =

$$\begin{pmatrix} 2e^t - e^{5t} \\ -2e^t + 3e^{5t} \end{pmatrix}$$

\]
</reasoning>

<answer>

Question 1

<mcq-option>If \(\mathbf{X}(0) = \mathbf{X}_0\) lies on the line \(y = -x\), then \(\mathbf{X}(t)\) approaches \((0, 0)\) along this line. Otherwise \(\mathbf{X}(t)\) approaches \((0, 0)\) from the direction determined by \(y = -3x\).</mcq-option>
<mcq-option>If \(\mathbf{X}(0) = \mathbf{X}_0\) lies on the line \(y = -3x\), then \(\mathbf{X}(t)\) approaches \((0, 0)\) along this line. Otherwise \(\mathbf{X}(t)\) approaches \((0, 0)\) from the direction determined by \(y = -x\).</mcq-option>
<mcq-correct>If \(\mathbf{X}(0) = \mathbf{X}_0\) lies on the line \(y = -x\), then \(\mathbf{X}(t)\) becomes unbounded along this line. Otherwise \(\mathbf{X}(t)\) becomes unbounded and \(y = -3x\) serves as an asymptote.</mcq-correct>
<mcq-option>If \(\mathbf{X}(0) = \mathbf{X}_0\) lies on the line \(y = -3x\), then \(\mathbf{X}(t)\) becomes unbounded along this line. Otherwise \(\mathbf{X}(t)\) becomes unbounded and \(y = -x\) serves as an asymptote.</mcq-option>
<mcq-option>All solutions spiral toward \((0, 0)\).</mcq-option>

Question 2

\[\mathbf{X}(t) =

$$\begin{pmatrix} 2e^t - e^{5t} \\ -2e^t + 3e^{5t} \end{pmatrix}$$

\]
</answer>

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