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PAGE 1

Direction Field / Slope Field (Continued)

Review: Linear Autonomous ODE

Last time: \( \frac{dy}{dx} = y \)

Figure: coordinate graph showing slope field and solution curves for \frac{dy}{dx} = y
Figure: Figure: coordinate graph showing slope field and solution curves for \( \frac{dy}{dx} = y \) Show Details
Coordinate slope field for the differential equation \( \frac{dy}{dx} = y \). Line segments represent local slopes \( y' \) depending exclusively on the vertical value \( y \). When \( y > 0 \), slopes are positive and increase with \( y \), yielding exponential growth curves. When \( y < 0 \), slopes are negative and direct curves downward into exponential decay. Along the \( x \)-axis (\( y = 0 \)), slopes are horizontal.

\( y' = y \) (height)

The slope at each point depends entirely on the vertical position (height \( y \)). As \( y \) increases, slopes become steeper and positive. When \( y < 0 \), slopes are negative.

Example: Nonlinear Autonomous ODE

Let's look at another one: \[ y' = y(y-2) \]

First, notice \( y' = 0 \) (\( \text{slope} = 0 \)) at \( y = 0 \) and \( y = 2 \).

\( y' = 0 \implies \text{equilibrium solutions} \)

Equilibrium solutions occur where the derivative is zero, meaning constant solutions \( y(x) = c \) that graph as horizontal lines.

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Slope Field and Qualitative Analysis for \( y' = y(y - 2) \)

Figure: slope field and solution curves partitioned across equilibrium values for y' = y(y-2)
Figure: Figure: slope field and solution curves partitioned across equilibrium values for \( y' = y(y-2) \) Show Details
Phase diagram and direction field for the autonomous differential equation \( y' = y(y-2) \). Horizontal dashed lines at \( y = 2 \) and \( y = 0 \) designate equilibrium lines with zero slope. In the top region (\( y > 2 \)), \( y' > 0 \) and slopes become steeper at higher values of \( y \), causing solution curves to grow toward \( +\infty \). In the middle region (\( 0 < y < 2 \)), \( y' < 0 \), directing trajectories downward toward \( y = 0 \). In the bottom region (\( y < 0 \)), \( y' > 0 \) with slopes steeper at lower heights, causing curves to rise asymptotically toward \( y = 0 \).

Qualitative Long-Term Behavior

Qualitatively, based on the initial condition \( y(0) \):

  • If the initial condition \( y(0) > 2 \),
    then as \( x \to \infty \), \( y \to \infty \).
  • If \( 0 < y(0) < 2 \),
    then as \( x \to \infty \), \( y \to 0 \).
  • If \( y(0) < 0 \),
    then as \( x \to \infty \), \( y \to 0 \).

We don't have the explicit formula for the solution \( y(x) \), but we understand its long-term behavior.

Stability Analysis:

The equilibrium solution \( y = 0 \) acts as an asymptotically stable equilibrium (sink/attractor) for initial values with \( y(0) < 2 \), since solutions approach \( 0 \) as \( x \to \infty \). Conversely, the equilibrium \( y=2 \) is an unstable equilibrium (source/repeller) because solutions near it move away as \( x \) increases.

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Qualitative Analysis: Introduction to Nullclines

Let's look at \( y' = -1 + t + y \) (\( t \) and \( x \) are common independent variables)

( If \( y' \) does not depend on the independent variable, then the diff. eq. is said to be autonomous )

Autonomous vs. Non-Autonomous ODEs: An ordinary differential equation is called autonomous if the independent variable (usually \( t \)) does not explicitly appear in the equation, i.e., \( y' = f(y) \). When \( t \) is present, as in \( y' = f(t, y) = -1 + t + y \), the equation is non-autonomous.

Finding the Nullcline

If independent variable is present, it's usually best to find the curve on which \( y' \) is \( 0 \):

→ nullclines: ( curve on which \( y' = 0 \) )

Applying this to our differential equation:

\[\begin{aligned}y' &= -1 + t + y \\[4pt]0 &= -1 + t + y \implies y = 1 - t \quad \text{nullcline}\end{aligned}\]
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Slope Fields and Solution Trajectories

Direction field and solution trajectories for differential equation y' = -1 + t + y with nullcline y = 1 - t
Figure: Direction field and solution trajectories for differential equation \( y' = -1 + t + y \) with nullcline \( y = 1 - t \) Show Details
Direction field and qualitative solution curves on the \( (t, y) \)-plane for \( y' = -1 + t + y = y - (1 - t) \). The nullcline is the straight line \( y = 1 - t \) along which \( y' = 0 \). In the region above the line (\( y > 1 - t \)), slope tangents are positive and trajectory paths curve upward. In the region below (\( y < 1 - t \)), slope tangents are negative and trajectory paths bend downward.

Above the Nullcline

\[\begin{aligned}y' &= -1 + t + y \\y' &= y - (1 - t)\end{aligned}\]

If \( y > 1 - t \) then \( y' > 0 \):

  • above nullcline, \( y' > 0 \)
  • more above, the steeper

Below the Nullcline

nullcline \( y = 1 - t \) (\( y' = 0 \))

\[y' = y - (1 - t)\]

If \( y < 1 - t \) (below nullcline) then \( y' < 0 \):

  • more below, steeper slopes

Qualitative Behavior: The nullcline \( y = 1 - t \) divides the phase plane into two regions. Solutions that start above the nullcline have positive derivative \( y' > 0 \) and curve upward toward \( +\infty \). Solutions below have negative derivative \( y' < 0 \) and curve downward toward \( -\infty \).

This gives us some qualitative understandings w/o solving the equation analytically

PAGE 5

First-Order Differential Equations: Physical Modeling

One equation we can already solve w/ calculus is:

\[ \frac{dy}{dt} = ay + b \quad (a, b \text{ constants}) \]
Concept Note: This is a standard first-order linear ordinary differential equation with constant coefficients, which can be solved using separation of variables or an integrating factor.

Simple but can be used to model many physical situations.

For Example, Free Fall

Consider a falling body subject to gravity and linear air resistance:

Free body diagram of a mass m falling under gravity and linear drag.
Figure: Free body diagram of a mass \( m \) falling under gravity and linear drag. Show Details
Free-body force diagram representing an object of mass \( m \). An upward force represents air resistance (drag) with magnitude \( \text{drag} = cv \), pointing upward opposite to motion, where \( c \) is the drag coefficient and \( v \) is velocity. A downward force represents gravitational force with magnitude \( \text{weight} = mg \), where \( g \) is acceleration due to gravity.
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Derivation of the Free Fall Equation

If \( y \) is the height, then velocity is \( v = \frac{dy}{dt} \).

\[ \text{Newton's 2nd Law: } F = m \underbrace{a}_{\text{accel.}} = m \frac{dv}{dt} \]

Summing the downward force of gravity and the upward drag force yields:

\[ m \frac{dv}{dt} = mg - cv \implies \frac{dv}{dt} = g - \frac{c}{m}v \]
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Solving by Substitution and Integration

Divide by \( y - \frac{5}{2} \) \( \left(y \neq \frac{5}{2}\right) \):

\[ \frac{1}{y - \frac{5}{2}} \frac{dy}{dt} = -2 \]
Dividing both sides by the expression involving \( y \) separates the dependent variable terms from the constant rate term.

Let \( u = y - \frac{5}{2} \), then \( \frac{du}{dt} = \frac{dy}{dt} \).

The equation above becomes:

\[ \frac{1}{u} \frac{du}{dt} = -2 \]

Integrating both sides with respect to \( t \):

\[ \int \frac{1}{u} \underbrace{\frac{du}{dt} \, dt}_{du} = \int -2 \, dt \]
Applying the chain rule substitution transforms the integral with respect to \( t \) into a direct integral with respect to \( u \).
\[ \int \frac{1}{u} \, du = \int -2 \, dt \] \[ \ln |u| = -2t + C \]
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General Solution Formulation

Exponentiating both sides to solve for \( |u| \):

\[ |u| = e^{-2t + C} = e^{-2t} \cdot e^C \]
\[ u = \underbrace{\pm e^C}_{\substack{\text{constant} \\ \text{call it } k}} \cdot e^{-2t} = k e^{-2t} \]
Recall: \( u = y - \frac{5}{2} \)
The factor \( \pm e^C \) combines the arbitrary integration constant and the sign from removing the absolute value into a single real constant \( k \).

So,

\[ y(t) = \frac{5}{2} - k e^{-2t} \]