Download original notes
PAGE 1

1.3 Slope Field and Solution Curves

A qualitative way to understand the solution.

From last time: \[ \frac{dy}{dx} = 2x \]

Solution: \( y = x^2 + c \)

Coordinate graph showing multiple parabolas shifted vertically, with short red tangent segments along the curves.

\( x^2 + 4 \)

\( x^2 + 2 \)

\( x^2 \)

\( x^2 - 2 \)

\( x^2 - 5 \)

Sketch lines tangent to curves at different points.

Each has slope given by the differential eq.

\[ \frac{dy}{dx} = 2x \]
PAGE 2

Pretend we don't know solution is \( x^2 + c \).

Construct a field of slopes using \[ \frac{dy}{dx} = 2x \]

A Cartesian plane showing a slope field with vertical columns of identical slopes and a red line through the origin.

redrawn see next page

\[ \frac{dy}{dx} = 2x \rightarrow \text{slope at point } (x, y) \]
  • \( (0, 0) \rightarrow \text{slope is } 2(0) = 0 \)
  • \( (1, 1) \rightarrow \text{slope is } 2 \)

Note it does not depend on \( y \): each column (fix \( x \)) has same slopes.

PAGE 3

Slope Fields and Solution Curves

\[ \frac{dy}{dx} = 2x \]

The solution curves can be seen in the slope field w/o having to solve for them.

We can see that as:

  • \( x \to \infty, y \to \infty \)
  • and \( x \to -\infty, y \to \infty \)
Slope field for dy/dx = 2x with red slope segments and green parabolic solution curves passing through the y-axis.
Figure 1: Slope field and solution curves for \( \frac{dy}{dx} = 2x \).

initial condition

PAGE 4

Example: \( y' = y \)

Slope field for y' = y with red segments and green exponential curves diverging from the horizontal axis.
Figure 2: Slope field and solution curves for \( y' = y \).

This time slope depends on \( y \) only. Each row all same slopes.

\( y = 0 : y' = 0 \)

Sketch a few solution curves using the slopes.

Even w/o solving, we see exponential behavior:

  • If \( y(0) > 0 \to y \to \infty \)
  • \( y(0) < 0 \to y \to -\infty \)
PAGE 5

Example: Differential Equation Analysis

\[ \frac{dy}{dx} = y - x \]

The slope at any point \( (x, y) \) is given by \( y' = y - x \).

Notice if \( y - x = 0 \implies \) on the curve \( y = x \), every point has \( y' = 0 \).

Hand-drawn slope field for dy/dx = y-x showing red slope segments and green solution curves.

Figure 1: Visual representation of the slope field and solution trajectories.

\( y = x \) (on here, \( \frac{dy}{dx} = y - x = 0 \))

Region Analysis

  • Above \( y = x \): then \( y - x > 0 \implies y' > 0 \)
  • Below \( y = x \): then \( y' < 0 \)

Asymptotic Behavior

We can say:

  • If \( y(0) \) is above \( y = x \), then \( y \to \infty \)
  • If \( y(0) \) is below \( y = x \), then \( y \to -\infty \)
PAGE 6

\[ \frac{dy}{dx} = y - x \]

A precise slope field for dy/dx = y-x with multiple green solution curves diverging from y=x.

Figure 2: Detailed slope field and integral curves for the differential equation.

PAGE 7

Slope Field Analysis

\[ \frac{dy}{dx} = y - \sin x \]

Slope Characteristics

  • On \( y = \sin x \), slope \( = 0 \)
  • Above \( y = \sin x \), slope increases
  • Below \( y = \sin x \), slope decreases

Visualizing the Solution

The following slope field illustrates the behavior of the differential equation \( \frac{dy}{dx} = y - \sin x \). Several solution curves are plotted in green, showing how they follow the direction of the slope field and behave relative to the sine curve.

Slope field for dy/dx = y - sin(x) with several green solution curves showing divergent and oscillatory behavior.
PAGE 8

Terminal Velocity Analysis

Let's use the slope field to analyze the terminal velocity of a falling object.

Physical Model

We consider an object falling under the influence of gravity and air resistance. In this model, we define the downward direction as positive velocity.

Free body diagram of a falling object with upward air resistance and downward gravity vectors.

Deriving the Differential Equation

According to Newton's 2nd Law:

\[ F = ma = m \frac{dv}{dt} \]

where \( v \) is velocity.

The net force is the difference between gravity and drag:

\[ F = \text{gravity} - \text{drag} \]

We can model drag as \( cv \), where \( c \) is a constant:

\[ F = mg - cv \]

Combining these expressions gives the final differential equation:

\[ m \frac{dv}{dt} = mg - cv \implies \frac{dv}{dt} = g - \frac{cv}{m} \]
PAGE 9

Differential Equations: Terminal Velocity

\[ \frac{dv}{dt} = g - \frac{cv}{m} \]

The slope is zero when: \[ g = \frac{cv}{m} \quad \text{or} \quad v = \frac{mg}{c} \]

Slope Field Analysis

  • Above \( v = \frac{mg}{c} \): slope is \( < 0 \). Higher up \( \rightarrow \) more negative.
  • Below \( v = \frac{mg}{c} \): opposite (slope is positive).
A slope field graph of velocity v versus time t, showing solution curves converging to a dashed horizontal line at v = mg/c.

Behavior of Solutions

  • If \( v(0) < \frac{mg}{c} \rightarrow v \) increases until \( v = \frac{mg}{c} \)
  • If \( v(0) > \frac{mg}{c} \rightarrow \) slow down and stay at \( v = \frac{mg}{c} \)

\( v = \frac{mg}{c} \) is the terminal velocity.

\( c \) is the drag coefficient (parachute \( \rightarrow \) high \( c \) so low \( \frac{mg}{c} \)).