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Exam 1 Review

exam format: 8 questions

"hybrid" multiple-choice

25% on answer
75% on work

4" x 6" note card

handwritten
no sharing

Defective system

\[ \vec{x}' = \begin{bmatrix} 4 & 1 \\ 0 & 4 \end{bmatrix} \vec{x} \]

\( \lambda = 4, 4 \)

\( (A - \lambda I) \vec{v} = \vec{0} \)

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\[ \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \quad \vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \text{missing one eigenvector} \]

generalize eigenvector \( \vec{u} \)

\[ (A - \lambda I) \vec{u} = \vec{v} \]

\[ \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \quad \vec{u} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

\[ \vec{x}(t) = c_1 e^{4t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{4t} \left( t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right) \]

Figure: A phase portrait diagram in the x1-x2 plane. The horizontal x1-axis has green arrows pointing away from the origin in both directions. Several blue curved trajectories originate near the origin and sweep outwards, eventually aligning with dashed red diagonal lines that serve as asymptotes. The vertical x2-axis is also shown. The origin is described by the adjacent text as an unstable nodal source.

origin:
unstable nodal source
improper
↓
follow asymptote

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\[ \vec{x}' = \begin{bmatrix} 1 & -2 \\ 5 & -1 \end{bmatrix} \vec{x} \]

\( \lambda = 3i, -3i \)
\( \vec{v} = \begin{bmatrix} 2 \\ 1-3i \end{bmatrix}, \begin{bmatrix} 2 \\ 1+3i \end{bmatrix} \)

Solution:

\[ e^{3it} \begin{bmatrix} 2 \\ 1-3i \end{bmatrix} = e^{i(3t)} \begin{bmatrix} 2 \\ 1-3i \end{bmatrix} \]\[ = (\cos(3t) + i \sin(3t)) \begin{bmatrix} 2 \\ 1-3i \end{bmatrix} \]\[ = \begin{bmatrix} 2 \cos(3t) + i 2 \sin(3t) \\ \cos(3t) + 3 \sin(3t) + i \sin(3t) - 3i \cos(3t) \end{bmatrix} \]\[ = \underbrace{\begin{bmatrix} 2 \cos(3t) \\ \cos(3t) + 3 \sin(3t) \end{bmatrix}}_{\vec{u}} + i \underbrace{\begin{bmatrix} 2 \sin(3t) \\ \sin(3t) - 3 \cos(3t) \end{bmatrix}}_{\vec{v}} \]

general solution: \( \vec{x} = c_1 \vec{u} + c_2 \vec{v} \)

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phase portrait: \(\lambda = \pm 3i\) Origin is center, solutions are ovals

orientation of ovals

\[ \begin{bmatrix} 2 \\ 1-3i \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} + i \begin{bmatrix} 0 \\ -3 \end{bmatrix} \]

Figure: A phase portrait diagram showing a set of axes with an elliptical trajectory centered at the origin. Arrows on the ellipse indicate a counter-clockwise flow. A green line segment is drawn through the origin, crossing the ellipse.

direction: \(\vec{x}' = \begin{bmatrix} 1 & -2 \\ 5 & -1 \end{bmatrix} \vec{x}\)

pick \(\vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\)

\(\vec{x}' = \begin{bmatrix} 1 \\ 5 \end{bmatrix}\)

up, right

\(x' = x(1-y)\)

\(y' = y(x-3)\)

Critical points: where \(x' = 0\) and \(y' = 0\)

\(x=0, y=1\) (from \(x' = 0\))

\(y=0, x=3\) (from \(y' = 0\))

cp: (0,0), (3,1)

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linearized system near cp \(\rightarrow\) Jacobian matrix

\[ x' = f(x, y) \]

\[ y' = g(x, y) \]

\[ J = \begin{bmatrix} f_x & f_y \\ g_x & g_y \end{bmatrix} \]

\[ x' = x - xy \]

\[ y' = -3y + xy \]

\[ J = \begin{bmatrix} 1 - y & -x \\ y & x - 3 \end{bmatrix} \]

\[ J(0, 0) = \begin{bmatrix} 1 & 0 \\ 0 & -3 \end{bmatrix} \quad \lambda = 1, -3 \quad \text{saddle pt, unstable} \]

\[ J(3, 1) = \begin{bmatrix} 0 & -3 \\ 1 & 0 \end{bmatrix} \quad \lambda = \text{imginary} \quad (3, 1) \text{ is a center, stable} \]

Competition:

\[ x' = x(1 - \frac{1}{2}x - y) \]

\[ y' = y(2 - \frac{1}{5}y - 2x) \]

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\[ f(t) = \begin{cases} 0 & 0 < t \le \pi \\ \sin(t) & \pi < t < \infty \end{cases} \]

\[ = u_{\pi}(t) \sin(t) \]

\[ F(s) = \mathcal{L} \{ u_{\pi}(t) \sin(t) \} \]

shift left: \( t \to t + \pi \)

\[ = e^{-\pi s} \mathcal{L} \{ \sin(t + \pi) \} \]

\( \sin(a+b) = \sin(a) \dots \)

\( \sin(t+\pi) = -\sin(t) \)

Figure: A hand-drawn graph showing a sine wave on a horizontal axis. The wave completes one full cycle from 0 to 2\pi. A dashed line indicates the continuation of the periodic function.

\[ = e^{-\pi s} \mathcal{L} \{ -\sin(t) \} \]

\[ = e^{-\pi s} \cdot \frac{-1}{s^2 + 1} \]

Convolution

\[ \int_{0}^{t} f(\tau) g(t-\tau) d\tau = \int_{0}^{t} f(t-\tau) g(\tau) d\tau = f * g \]

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\[ \mathcal{L}\{f * g\} = FG \]

for example, \[ \int_{0}^{t} (t-\tau)^2 \delta(\tau-3) d\tau \]

\[ f(t) = t^2 \quad g(t) = \delta(t-3) \]

Laplace transform is \[ \frac{2}{s^3} e^{-3s} \]

go back to \( t \): \[ \mathcal{L}^{-1} \left\{ e^{-3s} \frac{2}{s^3} \right\} \]

\[ = u_3(t) \cdot (t-3)^2 \]

\( t^2 \) shift RIGHT by 3