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Problem 22

\[ y' = \frac{-t + \sqrt{t^2 + 4y}}{2}, \quad y(2) = -1 \]

Solutions given:

\[ y_1 = 1 - t, \quad y_2 = -t^2/4 \]

b) Existence and Uniqueness

The initial value problem \( y' = f(t, y) \), \( y(t_0) = y_0 \) has a unique solution near \( (t_0, y_0) \) if \( f \) and \( \frac{\partial f}{\partial y} \) are both continuous.

\[ f = \frac{-t + \sqrt{t^2 + 4y}}{2} \text{ at } (2, -1) \]

\[ = \frac{-2 + \sqrt{4 - 4}}{2} = -1 \text{ is continuous} \]

\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left[ -\frac{1}{2}t + \frac{1}{2}(t^2 + 4y)^{1/2} \right] = \frac{1}{4}(t^2 + 4y)^{-1/2}(4) \]

\[ = \frac{1}{\sqrt{t^2 + 4y}} \quad \text{DNE at } (2, -1) \]

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Problem D

\[ \frac{dy}{dt} = y^2 - 4y, \quad y(0) = 8 \]

What is the largest interval where the solution is defined?

Separation of Variables

\[ \frac{dy}{dt} = y(y - 4) \]

\[ \frac{1}{y(y - 4)} dy = dt \]

Partial Fraction Decomposition

\[ \frac{1}{y(y - 4)} = \frac{A}{y} + \frac{B}{y - 4} \]

\[ 1 = A(y - 4) + B(y) \]

\[ 0y + 1 = (A + B)y - 4A \]

\( A + B = 0 \implies B = -A = \frac{1}{4} \)

\( 1 = -4A \implies A = -\frac{1}{4} \)

Integration

\[ \int \left( -\frac{1}{4} \frac{1}{y} + \frac{1}{4} \frac{1}{y - 4} \right) dy = \int dt \]

\[ -\frac{1}{4} \ln|y| + \frac{1}{4} \ln|y - 4| = t + k \]

\[ \ln|y| - \ln|y - 4| = -4t + a \]

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Solving the Differential Equation

\[ \ln \left| \frac{y}{y-4} \right| = -4t + a \]

Exponentiating both sides to solve for the ratio:

\[ \frac{y}{y-4} = e^{-4t+a} = Ce^{-4t} \]

Given the initial condition \( y(0) = 8 \):

\[ \frac{8}{4} = C = 2 \]

Substituting \( C = 2 \) back into the equation:

\[ \frac{y}{y-4} = 2e^{-4t} \]

Rearranging to solve for \( y \):

\[ y = 2e^{-4t}y - 8e^{-4t} \]\[ y - 2e^{-4t}y = -8e^{-4t} \]\[ y(1 - 2e^{-4t}) = -8e^{-4t} \]

The final explicit solution for \( y \) is:

\[ y = \frac{-8e^{-4t}}{1 - 2e^{-4t}} \]

Finding the Vertical Asymptote

To find where the solution is undefined, set the denominator to zero:

\[ 1 - 2e^{-4t} = 0 \]\[ 2e^{-4t} = 1 \]\[ e^{-4t} = \frac{1}{2} \]\[ t = \dots \]

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Differential Equation Analysis

DE:\( y' = y^2 - 4y \)\( y(0) = 8 \)

\( f(t, y) = y^2 - 4y \) is continuous for all \( y \)

\( \frac{\partial f}{\partial y} = 2y - 4 \) is continuous for all \( y \)

The continuity of the function and its partial derivative ensures the existence and uniqueness of a solution in a region around the initial point.

Figure: A coordinate system with t and y axes showing a point at (0, 8) enclosed in a rectangular region.

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2.5 Autonomous Equations

In the form

\[ \frac{dy}{dt} = f(y) \]

\( f \) does not depend on \( t \)

slope does not depend on \( t \)

e.g. population: \[ \frac{dy}{dt} = ry \]

always separable whether linear or nonlinear

When \[ \frac{dy}{dt} = 0 \rightarrow \text{equilibrium solutions or critical points} \]

Stability?

(do solutions converge to or diverge from them?)

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Example:

\[ \frac{dy}{dt} = y^2(y-2)(y-3) \]

Equilibrium solutions: \( \frac{dy}{dt} = 0 \)

\( y=0, \quad y=2, \quad y=3 \)

Check sign of \( \frac{dy}{dt} \) between/outside these:

\( y' = f(y) \)

Figure: Phase line with critical points at 0, 2, and 3. Signs are positive, positive, negative, positive respectively.

\( y' > 0 \)

\( y \) is increasing

moves to right

\( y' < 0 \)

\( y \) is decreasing

\( f(y) \)

Graph of \( f(y) \) vs. \( y \)

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\(y = 2\) is asymptotically stable because it attracts nearby solutions.
\(y = 3\) is unstable because solutions diverge from it.
\(y = 0\) is semistable because it is stable from one side, unstable from the other.

Phase Line and Solution Curves

The following diagrams illustrate the stability of the equilibrium solutions \(y=0\), \(y=2\), and \(y=3\) over time \(t\).

Figure: Vertical phase line with equilibrium points at 0, 2, and 3. Arrows show flow towards 2 and away from 3 and 0.

Figure: Coordinate graph of y versus t showing solution curves approaching y=2 and moving away from y=3 and y=0.

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Example: Gompertz Equation

\[ \frac{dy}{dt} = r y \ln\left(\frac{K}{y}\right), \quad y > 0 \]

\(r\): growth rate w/o limiting factors
\(K\): saturation level

\(\rightarrow\) used to model population in confined spaces (limited resources)

Growth rate is slow at beginning and end.

Critical Points

Critical points: \(y = 0, \quad y = K\)

Figure: Horizontal phase line with points 0 and K. Arrows indicate flow away from 0 and towards K.

Figure: Graph of f(y) vs y showing a parabolic curve starting at 0, peaking, and returning to the axis at K.

Stability

\(y = 0\) : unstable
\(y = K\) : asymp. stable

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Logistic Growth Visualization

This graph illustrates the behavior of a population or quantity following a logistic growth model over time.

Growth Curve and Carrying Capacity

The graph displays the relationship between time \( t \) and the quantity \( y \). The curve starts at an initial value on the vertical axis and increases sigmoidally, approaching a horizontal asymptote. This asymptote represents the carrying capacity, labeled here as \( K \).

As \( t \to \infty \), the value of \( y \) approaches the limit \( K \), indicating a stable equilibrium state.

Figure: A graph showing a sigmoidal curve starting on the y-axis and approaching a horizontal dashed line at y=K as t increases.