Field Flows Slides Visualizing Flow
Direction Fields & Existence Theorems
Analyzing ODEs without the Algebra
The Particle's Path
Imagine a particle dropped into a stream.
We don't know the exact position equation, but at every point $(x, y)$, we know the velocity vector .
How can we trace the path?
Does every start point have a path?
Can paths ever cross?
dy/dx = f(x,y)
Defining the Direction Field
\[ \frac{dy}{dx} = f(x, y) \]
At every point $(x, y)$ in the plane, the derivative gives us the slope of the solution curve passing through that point.
Step 1
Pick a point $(x, y)$
Step 2
Calculate $f(x, y)$
Step 3
Draw a line with that slope
The Power of Isoclines
Technique
Plotting point-by-point is tedious. Instead, look for curves where the slope is constant .
Isocline Equation:
\[ f(x, y) = c \]
Example: $\frac{dy}{dx} = x + y$
For $c = 0$, slope 0 on line $y = -x$
For $c = 1$, slope 1 on line $y = -x + 1$
For $c = -1$, slope -1 on line $y = -x - 1$
Isocline: f(x,y)=c
/
/
/
Existence and Uniqueness
Picard-Lindelöf Theorem
Consider the initial value problem: $\frac{dy}{dx} = f(x, y)$, with $y(x_0) = y_0$.
1. Existence
$f(x, y)$ is continuous on a region containing $(x_0, y_0)$.
2. Uniqueness
$\frac{\partial f}{\partial y}$ is continuous on that same region.
"If these hold, a unique solution exists near $(x_0, y_0)$ and trajectories can never cross."
Qualitative Toolkit
Direction Fields
Global visual map of solution behaviors.
Equilibrium
Where $dy/dx = 0$. Critical for stability.
Sound Models
Theorems ensure solutions exist and are unique.
Lab Session
Open the Slope Scout Worksheet .
Slope Scout Worksheet Slope Scout
Qualitative Analysis Worksheet
Name:
Date:
Part 1: Isocline Mapping
For the differential equation \(\frac{dy}{dx} = x + y\) , find the isoclines for the given slopes \(c\). Draw the isocline (dashed line) and sketch the slope marks along it.
1.1 Complete the Table
Slope \(c\) Isocline Equation \(0\) y = -x \(1\) \(-1\) \(2\)
x y
Sketch the direction field for \(\frac{dy}{dx} = x + y\)
Part 2: Existence and Uniqueness
Determine if the Picard-Lindelöf Theorem guarantees a unique solution for the following Initial Value Problems (IVPs) near the given point.
\[ \frac{dy}{dx} = \sqrt{y}, \quad y(0) = 0 \]
Check Continuity of \(f(x, y)\):
Check Continuity of \(\frac{\partial f}{\partial y}\):
Does a unique solution exist?
Yes
No
\[ \frac{dy}{dx} = x \ln(y), \quad y(1) = e \]
Check Continuity of \(f(x, y)\):
Check Continuity of \(\frac{\partial f}{\partial y}\):
Does a unique solution exist?
Yes
No
Part 3: Qualitative Prediction
A cooling body is modeled by \(\frac{dT}{dt} = -0.1(T - 20)\).
Without solving the equation, what is the equilibrium solution ? What happens to the temperature as \(t \to \infty\) if \(T(0) = 100\)?
Split and Solve Slides Split and Solve
Separation of Variables
Algebraic Keys to Dynamic Models
The Separable Form
A first-order ODE is separable if the derivative can be expressed as a product of a function of $x$ and a function of $y$.
\[ \frac{dy}{dx} = g(x) \cdot h(y) \]
The Strategy:
1 Move all $y$ terms with $dy$ and all $x$ terms with $dx$.
2 Integrate both sides.
Example:
\[ \frac{dy}{dx} = \frac{x^2}{y} \]
\[ y \, dy = x^2 \, dx \]
\[ \int y \, dy = \int x^2 \, dx \]
\[ \frac{1}{2}y^2 = \frac{1}{3}x^3 + C \]
Newton's Law of Cooling
The rate of change of temperature is proportional to the difference between the object and its surroundings.
\[ \frac{dT}{dt} = -k(T - T_s) \]
$T$
Object Temp
$T_s$
Ambient Temp
Separating leads to the solution:
\[ T(t) = T_s + (T_0 - T_s)e^{-kt} \]
Radioactive Decay
The amount of substance $N$ decays at a rate proportional to how much is left.
\[ \frac{dN}{dt} = -kN \]
Half-Life
The time required for the quantity to reduce by half.
\[ t_{1/2} = \frac{\ln(2)}{k} \]
Case File: 01-18-26
The Scene:
A body is found in a climate-controlled room at 10:00 PM. Room temperature is a constant $70^\circ\text{F}$.
Temp @ 10:00 PM: $80^\circ\text{F}$
Temp @ 11:00 PM: $75^\circ\text{F}$
When did death occur?
(Assume normal body temp was $98.6^\circ\text{F}$)
START INVESTIGATION
Cold Case Worksheet Cold Case Report
Forensic Mathematics Lab
Lead Investigator:
Case Date:
Evidence Log: Case 01-18-26
Ambient Temperature (\(T_s\)): \(70^\circ\text{F}\) (constant)Initial Body Temp (\(T_0\)): \(98.6^\circ\text{F}\) (estimated)First Reading (10:00 PM): \(80^\circ\text{F}\)Second Reading (11:00 PM): \(75^\circ\text{F}\)
Task 1: Model Derivation
Use Newton's Law of Cooling: \(\frac{dT}{dt} = -k(T - T_s)\). Separate the variables and integrate to find the general solution for \(T(t)\).
Show your separation of variables and integration steps below:
Task 2: Finding \(k\) and Time of Death
Let \(t = 0\) be 10:00 PM. Use the data points from the evidence log to calculate the cooling constant \(k\) and then determine how many hours before 10:00 PM the temperature was \(98.6^\circ\text{F}\).
Step A: Solve for \(k\) using readings at \(t=0\) and \(t=1\):
Step B: Solve for \(t_{death}\) where \(T(t_{death}) = 98.6^\circ\text{F}\):
Task 3: Rapid Separation Check
Solve the following separable initial value problem:
\[ \frac{dy}{dx} = \frac{x^2}{y^2}, \quad y(0) = 2 \]
Factor Finding Slides Factor Finding
Integrating Factors for Linear ODEs
Unlocking Non-Separable Systems
Standard Linear Form
Not every equation can be separated. A first-order linear ODE must first be written in this specific form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
Note: The coefficient of \(dy/dx\) must be 1 .
Why is this hard?
The left side looks almost like a product rule result... but it's missing a piece.
The Solution:
We multiply the entire equation by a special function \(\mu(x)\) to force the product rule.
The Integrating Factor
The Formula
\[ \mu(x) = e^{\int P(x) dx} \]
Step 1
Find \(P(x)\) from standard form.
Step 2
Compute \(\int P(x) dx\).
Step 3
Raise \(e\) to that power.
Why it works
After multiplying by \(\mu(x)\), the left side becomes the derivative of a product:
\[ \frac{d}{dx} \left[ \mu(x) y \right] = \mu(x) Q(x) \]
Now, we just integrate both sides with respect to \(x\) and solve for \(y\). No complex substitution required!
The Variable Inflow Tank
A tank contains 100L of water. Salt flows in at a rate that increases over time: \(r_{in} = 2t\) grams/min.
\[ \frac{dS}{dt} + \frac{1}{100}S = 2t \]
Try to separate this. You can't. You need an Integrating Factor.
Can you find \(\mu(t)\)?
Find the amount of salt \(S(t)\) at any time \(t\).
Linear Equation Checklist
Write in Standard Form
Identify \(P(x)\)
Calculate \(\mu(x)\)
Integrate \(\mu(x)Q(x)\)
"The Integrating Factor is the bridge between a non-separable mess and a solvable product rule."
Integrating Factor Worksheet Integrating Factor Lab
Solving Linear First-Order Equations
Name:
Date:
Part 1: The Integration Engine
For each equation, transform it into standard linear form , identify \(P(x)\), calculate the integrating factor \(\mu(x)\), and solve for the general solution.
1.1 \[ x \frac{dy}{dx} + 2y = \frac{\sin(x)}{x} \]
Standard Form & \(P(x)\):
Integrating Factor \(\mu(x)\):
General Solution Calculation:
1.2 \[ \frac{dy}{dt} - 2ty = t^3 \]
Integrating Factor \(\mu(t)\):
General Solution:
Part 2: The Variable Inflow Tank
A 200L tank is initially full of pure water. Saltwater is pumped in at 4 L/min with a concentration that varies as \(c_{in}(t) = 5 + 2\cos(t)\) grams/L. The well-mixed solution is pumped out at the same rate.
Set up and solve the IVP to find the amount of salt \(S(t)\) in the tank at any time.
ODE Formulation (Standard Form):
Calculating \(\mu(t)\):
Integrating and Applying Initial Condition \(S(0) = 0\):
Growth and Limits Slides Growth and Limits
Population Dynamics
From Infinite Skies to Crowded Islands
The Reindeer Paradox
In 1944, 29 reindeer were introduced to St. Matthew Island. With no predators and plenty of lichen, the population exploded.
1944: 29 reindeer
1963: 6,000 reindeer
1966: 42 reindeer (The Crash)
Why did the simple exponential model fail so catastrophically?
The Reality (Logistic)
The Crash
Unrestricted Growth
The Malthusian Model assumes growth is proportional only to current size.
\[ \frac{dP}{dt} = rP \]
Solution: \( P(t) = P_0 e^{rt} \)
Problem: It assumes infinite resources. In the real world, ecosystems have a maximum capacity.
Density Dependence
Verhulst Model
We modify the growth rate to include a term that slows down as the population approaches the Carrying Capacity (\(K\)) .
\[ \frac{dP}{dt} = rP \left( 1 - \frac{P}{K} \right) \]
If \(P \ll K\):
Growth is roughly exponential.
If \(P \to K\):
Growth rate \(dP/dt \to 0\).
Solving the Logistic ODE
This is a separable equation, but the integration requires partial fraction decomposition.
\[ \int \frac{1}{P(1 - P/K)} dP = \int r dt \]
Integrating this gives the general solution:
\[ P(t) = \frac{K}{1 + Ae^{-rt}} \]
where \(A = (K - P_0) / P_0\)
Stability Analysis
Setting \(dP/dt = 0\) reveals the steady states of the system.
\(P = 0\)
Unstable Equilibrium
Any small population will grow away from zero.
\(P = K\)
Stable Equilibrium
The population naturally trends toward the carrying capacity.
Population Growth Lab Growth and Limits Lab
Analyzing Population Dynamics
Name:
Date:
Part 1: The Malthusian Failure
In 1944, 29 reindeer were introduced to St. Matthew Island. By 1963, the population was 6,000.
1.1 Calculate the intrinsic growth rate \(r\):
Assume exponential growth \(P(t) = P_0 e^{rt}\). Let \(t=0\) be 1944 and \(t=19\) be 1963.
1.2 Predict the population in 1966 using this exponential model:
Actual Data:
In 1966, researchers found only 42 reindeer left on the island. The population had crashed due to overgrazing and a severe winter.
Part 2: Modeling with Limits
The logistic equation is given by: \(\frac{dP}{dt} = rP \left( 1 - \frac{P}{K} \right)\) .
2.1 Use partial fraction decomposition to show that:
\[ \int \frac{1}{P(1 - P/K)} dP = \ln \left| \frac{P}{1 - P/K} \right| + C \]
2.2 Stability Analysis
Sketch the phase line for the logistic equation. Label the equilibrium points and indicate with arrows the direction of the flow.
P = 0
P = K
Which equilibrium is stable ? Explain in terms of biological carrying capacity.
Part 3: Numerical Application
A fish population in a lake is modeled by \(\frac{dP}{dt} = 0.5P(1 - P/2000)\). The initial population is 500 fish.
Find the specific solution \(P(t)\) and calculate the population after 5 years.
Tanks and Circuits Slides Tanks and Circuits
Compartmental Modeling
Applications of First-Order Systems
Salt in a Tank
The rate of change of a substance is the Rate In minus the Rate Out .
\[ \frac{dS}{dt} = \text{rate}_{\text{in}} - \text{rate}_{\text{out}} \]
$\text{Rate} = \text{Flow Rate} \times \text{Concentration}$
Inflow
Outflow
Electrical Circuits
Using Kirchhoff's Voltage Law for a Resistor (\(R\)) and Capacitor (\(C\)) in series:
\[ R \frac{dq}{dt} + \frac{1}{C}q = E(t) \]
CHARGE
\(q\)
CURRENT
\(i = \frac{dq}{dt}\)
Key Insights
Long-term Behavior
Transient Solution
The part of the solution that vanishes as \(t \to \infty\). Usually the term associated with the complementary solution (\(e^{-t/\tau}\)).
Steady State
The part of the solution that remains as \(t \to \infty\). Usually driven by the external source \(E(t)\) or \(Q(t)\).
The Pollution Flush
A lake contains 1,000,000 gallons of polluted water. Pure water flows in at 50,000 gal/day.
Find the time to remove 90% of the pollutant.
What happens if the inflow is also slightly polluted?
Model the System
Apply the skills from the entire unit to solve this complex problem.
System Stress Worksheet System Stress Test
Applied Compartmental Analysis
Team Members:
Date:
Task 1: Lake Pollution Control
A small lake has a volume of 1,000,000 gallons . It is contaminated with 500 lbs of a specific pollutant. Fresh water flows into the lake at 20,000 gal/day , and the well-mixed mixture flows out at the same rate.
1.1 Formulate the Initial Value Problem (IVP):
Define your variables and set up the differential equation for the amount of pollutant \(A(t)\).
1.2 Solve for \(A(t)\):
1.3 Environmental Compliance:
The lake is considered "safe" when the pollutant level drops below 50 lbs. How many days will this take? Show your work.
Task 2: RC Circuit Response
A circuit contains a resistor (\(R = 100 \, \Omega\)) and a capacitor (\(C = 0.01 \, \text{F}\)) in series with a 12V battery. The switch is closed at \(t = 0\) with 0 initial charge on the capacitor.
2.1 Use Kirchhoff's Law to write the ODE for charge \(q(t)\):
\[ R \frac{dq}{dt} + \frac{1}{C}q = E(t) \]
2.2 Solve the IVP for \(q(t)\) using Integrating Factors:
2.3 Find the Transient term:
2.4 Find the Steady-State charge:
Part 3: Comparative Analysis
How is the "Lake Pollution" problem mathematically equivalent to the "RC Circuit" charging problem? Compare their structures and their behavior as \(t \to \infty\).
Instructor Solutions Manual Instructor Solutions Manual
Analyzing Dynamic Systems Sequence
Teacher Resource
Lesson 1: Slope Scout Worksheet
1.1 Isocline Table (\(dy/dx = x + y\)):
\(c=0 \implies y = -x\) (Slopes are horizontal)
\(c=1 \implies y = -x + 1\) (Slopes are \(45^\circ\))
\(c=-1 \implies y = -x - 1\) (Slopes are \(-45^\circ\))
\(c=2 \implies y = -x + 2\)
2.1 Existence & Uniqueness (\(dy/dx = \sqrt{y}, y(0)=0\)):
\(f(x,y) = \sqrt{y}\) is continuous for \(y \ge 0\). However, \(\partial f/\partial y = 1/(2\sqrt{y})\) is not continuous at \(y=0\). Thus, uniqueness is not guaranteed (solutions are \(y=0\) and \(y=x^2/4\)).
Lesson 2: Cold Case Worksheet
Forensic Calculation Summary:
1. \(T(t) = 70 + (T_0 - 70)e^{-kt}\). With \(T(0)=80\), we get \(80 = 70 + (T_0 - 70)e^0 \implies T_0-70 = 10\). So \(T(t) = 70 + 10e^{-kt}\).
2. At 11:00 PM (\(t=1\)), \(T(1)=75 \implies 75 = 70 + 10e^{-k} \implies 0.5 = e^{-k} \implies k = \ln(2) \approx 0.693\).
3. Set \(T(t) = 98.6 \implies 98.6 = 70 + 10e^{-0.693t} \implies 2.86 = e^{-0.693t} \implies t \approx -1.51\) hours.
Time of Death: Approximately 8:29 PM.
Lesson 3: Integrating Factor Worksheet
1.1 Solution (\(x dy/dx + 2y = \sin(x)/x\)):
Standard Form: \(y' + (2/x)y = \sin(x)/x^2\). \(P(x) = 2/x\).
\(\mu(x) = e^{\int 2/x dx} = e^{2\ln x} = x^2\).
Multiply: \(x^2 y' + 2xy = \sin(x) \implies [x^2 y]' = \sin(x)\).
Integrate: \(x^2 y = -\cos(x) + C \implies y = \frac{C - \cos(x)}{x^2}\).
Lesson 4: Population Growth Lab
1.1 Growth Rate (\(r\)):
\(6000 = 29e^{19r} \implies 206.9 = e^{19r} \implies r = \ln(206.9)/19 \approx 0.2806\) (28% annual growth).
2.2 Stability Analysis:
\(P=0\) is unstable (source); \(P=K\) is stable (sink). Biologically, \(K\) is stable because negative feedback (lack of food) reduces growth when \(P > K\), and high resources increase growth when \(P < K\).
Lesson 5: System Stress Worksheet
1.1 Lake Pollution IVP:
\(dA/dt = (\text{Rate In}) - (\text{Rate Out}) = 0 - (20,000 / 1,000,000)A = -0.02A\).
Solution: \(A(t) = 500e^{-0.02t}\). Set \(50 = 500e^{-0.02t} \implies 0.1 = e^{-0.02t} \implies t = \ln(10)/0.02 \approx 115\) days.
2.2 Circuit Response:
\(100 q' + 100 q = 12 \implies q' + q = 0.12\). \(\mu(t) = e^t\).
\(q(t) = 0.12(1 - e^{-t})\). Steady state charge: \(0.12 \, \text{C}\).
