100%
Download original notes
PAGE 1

Some Basic Mathematical Models and Direction Field

differential equation: any equation containing derivative(s)

Examples from Calculus

you have seen some: for example,

\[ \frac{dy}{dx} = e^x \]

or

\[ y' = x^2 + 2x - 5 \]
from calculus

of course, they get more complicated:

\[ y'' + 3y' + y = 2 \]
models a mass-spring-damper system with constant external force
Physics Context: In a second-order linear ODE like \( y'' + 3y' + y = 2 \), \( y'' \) represents acceleration (inertial mass term), \( 3y' \) is the damping friction term proportional to velocity, \( y \) is the restoring spring force proportional to displacement, and \( 2 \) represents a constant external applied force.

we use diff. eqs to model situations we want to study

PAGE 2

Basic Growth Models

\[ \underbrace{\frac{dy}{dt}}_{\substack{\text{rate of} \\ \text{change of } y}} = \underbrace{k}_{\substack{\text{constant of proportionality} \\ \text{("is proportional to")}}} \underbrace{y}_{y \text{ itself}} \]

exponential growth (\( k > 0 \))

"   decay (\( k < 0 \))

or

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

logistic growth

\( L \): carrying capacity (limit)

Key Model Insight: In \( \frac{dy}{dt} = ky \), the growth rate is directly proportional to the current quantity \( y \). In \( \frac{dy}{dt} = k(L-y) \), the rate of change is proportional to the remaining capacity \( L-y \) until the carrying capacity \( L \) is reached.

goal : solve the diff. eq.

find a function that satisfies the diff. eq.

for example,

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

find \( y(x) \) that satisfies the diff. eq.

\[ y(x) = \sin(x) + C \]

because

\[ \frac{d}{dx} (\sin x + C) = \cos x \]
PAGE 3

Differential Equations & Guessing Solutions

but \( y'' + 3y' + 2y = 5 \) is harder to solve (we'll learn how in this course)

we will learn many techniques

but sometimes we can “guess” a solution

for example,

\( y' = y \)

we cannot simply integrate both sides

\[ \begin{aligned} \int y' \, dx &= \int y \, dx \\[8pt] y &= \underbrace{\int y \, dx}_{?} \end{aligned} \]
Why Direct Integration Fails: We cannot evaluate \( \int y \, dx \) directly because \( y \) is an unknown function of \( x \), not a known integrand.

\( y' = y \longrightarrow \) says to find \( y \) such that it is its own derivative

  • \( y = e^x \)
  • so is \( y = 2e^x \) \( (y' = 2e^x) \)
  • so is \( y = 3e^x, \; 10e^x, \; \pi e^x \), etc.
PAGE 4

Family of Solutions & Geometric Meaning

\[ y = C e^x \]

\( C \): constant (\( C \) can be zero)

graph:

Graph showing a family of exponential curves y = C e^x for different constant values
Figure: Graph showing a family of exponential curves \( y = C \) \( e^x \) for different constant values Show Details
Coordinate graph displaying the family of solutions \( y = C e^x \) on Cartesian axes \( (x, y) \). The curves shown include \( y = 10e^x \) with \( y \)-intercept at \( (0, 10) \), \( y = e^x \) with \( y \)-intercept at \( (0, 1) \), and \( y = -3e^x \) with \( y \)-intercept at \( (0, -3) \).

solution is often a family of curves

\( y' = y \longrightarrow \) says that on the curve \( y(x) \) the slope (\( y' \)) is always equal to its \( y \)-value

Coordinate diagrams showing tangent slopes matching function height on y = e^x and reconstruction of solution
Figure: Coordinate diagrams showing tangent slopes matching function height on y = \( e^x \) and reconstruction of solution Show Details
Two side-by-side diagrams illustrating the geometric relationship for \( y' = y \). The left diagram shows the curve \( y = e^x \) where at height \( y = 1 \) the tangent has \( \text{slope } 1 \), at \( y = 3 \) the tangent has \( \text{slope } 3 \), and at \( y = 5 \) the tangent has \( \text{slope } 5 \). The right diagram shows dashed tangent segments of varying slopes reconstructing the solution curve.

we can “eyeball” solution from slopes

Geometric Interpretation: Since the differential equation dictates that the derivative \( y' \) equals the value \( y \), at any horizontal level \( y = k \), the tangent lines along all solution curves must have slope equal to \( k \).
PAGE 5

Slope Field (Direction Field) for \( y' = y \)

Let's construct a slope field (direction field) for \( y' = y \).

Figure: Slope field for the differential equation y' = y showing tangent segments and exponential solution curves.
Figure: Figure: Slope field for the differential equation \( y' = y \) showing tangent segments and exponential solution curves. Show Details
Slope field plotted on the \( (x, y) \)-plane for the differential equation \( y' = y \). Short red slope ticks indicate the derivative at various coordinates: horizontal segments (slope \( 0 \)) along the \( x \)-axis where \( y = 0 \), positive slopes of value \( 1 \) along \( y = 1 \), steeper positive slopes for \( y > 1 \), and negative slopes for \( y < 0 \) (such as slope \( -1 \) along \( y = -1 \)). Two dashed black curves illustrate particular solution trajectories following the direction field, behaving as exponential functions \( y = C e^x \).

Calculating Slopes

\( y' = y \)

  • If \( y = 1 \) (horizontal line): slopes at every \( x \) is \( 1 \)
  • If \( y = 0 \): \( \text{slopes} = 0 \)

Concept Breakdown: In an autonomous first-order ODE \( y' = f(y) \), the derivative \( y' \) depends solely on \( y \) and not on \( x \). This means along any horizontal line \( y = c \), every tangent segment has the exact same slope \( m = c \). As \( y \) increases, the slopes become steeper; when \( y < 0 \), the slopes are negative.

The idea: Use the slopes to visualize solutions qualitatively.