Differential Equations
Modeling change: first-order equations, oscillations and resonance, Laplace transforms, systems and phase planes, stability, numerical methods and the heat equation, with four computer labs on epidemics, predators and prey, climate and wind-turbine vibration.
Courses › MATH 220
About this course
This is a free self-study course outline written by Saving Planets. It is not a university course, it carries no academic credit, and no school is involved.
Its scope follows a sophomore course in ordinary differential equations, modeled on the topics and depth of MIT OpenCourseWare 18.03 and the free textbooks by Jiří Lebl (Notes on Diffy Qs) and William Trench, and on the syllabi of courses such as Georgia Tech MATH 2552 and Stanford Math 53. We link to the free materials throughout. None of those authors or organizations endorses this page.
What we add: every unit connects to a real planet problem with a cited public source (lakes, methane, fisheries, carbon, climate, epidemics, wind turbines), every number on the page is re-computed by a checking script before it goes live, and four computer labs solve real models numerically.
Work through it in order. Each unit has goals, key ideas, equations, step-by-step examples, practice with full answers, common mistakes, flashcards and a quiz. Your progress is saved only in this browser.
How this course helps save the planet
Almost every model of the living planet is a differential equation: how fast a lake flushes out a pollutant, how a fish stock recovers, how carbon moves between air and sea, how quickly Earth warms when its energy balance shifts, how an epidemic spreads, and how a wind-turbine tower sways.
Each unit ends with a short, sourced section showing how its ideas help both people and nature. Model parameters that are not from a source are labeled as sample values.
Before you start
Prerequisites
- Calculus I and II: derivatives, integrals, partial fractions, Taylor series.
- Matrices, determinants and eigenvalues of 2 × 2 matrices (taught briefly in Unit 6).
- Calculus III helps for the heat equation but is not required.
How to study
- Check every solution by substituting it back into the equation.
- Do every practice problem on paper before opening the answer.
- Sketch a slope field or phase portrait before solving; know what the answer should look like.
- Ask of every model: what are the units, and does the long-run behavior make sense?
- Solve one problem from each unit numerically too, and compare with the formula.
- Retake the quizzes a few days later. Spaced practice beats one long session.
Lab equipment (for the wet labs)
- A computer or tablet with a spreadsheet.
- A free graphing calculator such as Desmos.
- Optional: Python with NumPy (free in Google Colab).
Suggested 15-week schedule
About 6 to 9 hours a week: reading, practice problems and, in lab weeks, the lab.
| Week | Unit | Topics | Reading |
|---|---|---|---|
| 1 | First-Order Equations | Slope fields; separable equations | Lebl 1.1–1.3 |
| 2 | First-Order Equations | Linear and exact equations; existence and uniqueness | Lebl 1.2, 1.4, 1.8 |
| 3 | Modeling with First-Order Equations | Modeling: mixing, decay, logistic growth, harvesting | Lebl 1.6 |
| 4 | Second-Order Linear Equations and Oscillations | Second-order linear equations; characteristic equation | Lebl 2.1–2.2 |
| 5 | Second-Order Linear Equations and Oscillations | Mechanical vibrations and damping | Lebl 2.4 |
| 6 | Forced Vibration and Resonance | Undetermined coefficients; forced oscillation and resonance; Lab D | Lebl 2.5–2.6 |
| 7 | Laplace Transforms | Laplace transforms; solving IVPs | Lebl 6.1–6.2 |
| 8 | Laplace Transforms | Step functions, impulses, convolution; midterm practice | Lebl 6.3–6.4 |
| 9 | Linear Systems, Eigenvalues and Phase Planes | Systems; matrices and eigenvalues | Lebl 3.1–3.4 |
| 10 | Linear Systems, Eigenvalues and Phase Planes | Phase portraits and classification | Lebl 3.5 |
| 11 | Nonlinear Systems and Stability | Nonlinear systems, linearization; predator-prey; Lab B | Lebl 8.1–8.3 |
| 12 | Nonlinear Systems and Stability | Stability; SIR epidemics; Lab A | Lebl 8.2–8.3 |
| 13 | Numerical Methods | Numerical methods; energy-balance model; Lab C | Lebl 1.7, Trench 3 |
| 14 | Series Solutions, Fourier Series and the Heat Equation | Series solutions; Fourier series; heat equation | Lebl 7.1–7.2, 4.2, 4.6 |
| 15 | Review | Review and final practice set | All units |
Your progress
Tick units and labs as you finish them, and take the quizzes. Progress and quiz scores are saved only in this browser, never sent to us.
Turn on JavaScript to track progress.
Unit 1 · Weeks 1–2
First-Order Equations
Slope fields, separable equations, linear equations with integrating factors, exact equations, and when a solution exists and is unique.
Learning objectives
By the end of this unit you can:
- Read a slope field and sketch solutions.
- Solve separable and linear first-order equations.
- Recognize and solve exact equations.
- State the existence and uniqueness theorem and spot when it fails.
Key concepts
- A differential equation relates an unknown function to its derivatives; a solution is a function that makes it true.
- Separable: dy/dx = g(x)h(y). Divide and integrate both sides.
- Linear: y′ + p(x)y = q(x). Multiply by the integrating factor μ = e^(∫p dx) so the left side becomes (μy)′.
- Exact: M dx + N dy = 0 with M_y = N_x is dF = 0, so F(x, y) = C.
- If f and ∂f/∂y are continuous near (x₀, y₀), then y′ = f(x, y), y(x₀) = y₀ has exactly one solution nearby. Solutions may still blow up in finite time.
Key equations
Worked examples, step by step
Example 1.1 Solve dy/dx = xy, y(0) = 2, and find y(1).
Show the steps
- dy/y = x dx, so ln|y| = x²/2 + C.
- y = 2e^(x²/2).
- y(1) = 2e^0.5.
Answer: ≈ 3.297.
Example 1.2 Solve y′ + 2y = 4, y(0) = 0, and find y(1).
Show the steps
- μ = e^(2t), so (e^(2t)y)′ = 4e^(2t).
- e^(2t)y = 2e^(2t) + C; y(0) = 0 gives C = −2.
- y = 2(1 − e^(−2t)).
Answer: y(1) ≈ 1.729, approaching the steady state 2.
Example 1.3 Solve 2xy dx + (x² + 2y) dy = 0 with y(0) = 1, and find y when x = 1.
Show the steps
- M_y = 2x = N_x: exact.
- F = x²y + y²; F(0, 1) = 1.
- At x = 1: y + y² = 1, so y = (−1 + √5)/2.
Answer: ≈ 0.618.
Practice problems
Work each on paper first, then check.
Solve y′ = y², y(0) = 1. Where does the solution exist?
Show answer
y = 1/(1 − t); it blows up at t = 1, even though y² is smooth everywhere.
Solve t y′ + y = t² for t > 0.
Show answer
(ty)′ = t², so y = t²/3 + C/t.
Solve y′ = −y/τ.
Show answer
y = y₀e^(−t/τ).
Is (y cos x) dx + (sin x + 2y) dy = 0 exact? Solve.
Show answer
M_y = cos x = N_x; F = y sin x + y² = C.
Does y′ = √y, y(0) = 0 have a unique solution?
Show answer
No: y = 0 and y = t²/4 both work; ∂f/∂y = 1/(2√y) is not continuous at y = 0.
Common misconceptions
Myth: Every first-order equation is separable.
Actually: y′ = x + y is not; it is linear.
Myth: Forgetting +C until the end is fine.
Actually: Add C when you integrate, before solving for y, or the initial condition comes out wrong.
Myth: A smooth right-hand side means the solution lives forever.
Actually: y′ = y² blows up at a finite time.
Flashcards
Separable
dy/dx = g(x)h(y).
Linear first-order
y′ + p y = q.
Integrating factor
μ = e^(∫p dx).
Exact
M_y = N_x; F = C.
Initial value problem
An ODE plus y(x₀) = y₀.
Existence and uniqueness
f, f_y continuous ⇒ one local solution.
Self-check quiz
How this helps save the planet
The simplest first-order equation, y′ = −y/τ, describes how a pollutant clears from a well-mixed lake or how a gas leaves the atmosphere. Its one number, the time scale τ, decides whether a mistake fades in years or lasts for centuries (Unit 2).
On Saving Planets: Rivers hub
Sources: U.S. EPA: Physical features of the Great Lakes (retention times) · Lebl, 1.4 Linear equations and the integrating factor
Free learning resources
- Lebl, 1.3 Separable equations: Free textbook section.
- Lebl, 1.4 Integrating factor: Linear equations.
- Lebl, 1.8 Exact equations: Exact equations.
- Lebl, 1.2 Slope fields: Existence and uniqueness.
Unit 2 · Week 3
Modeling with First-Order Equations
Turning a situation into an equation: mixing and lake pollution, decay and lifetimes, Newton cooling, logistic growth, harvesting, and equilibria with their stability.
Learning objectives
By the end of this unit you can:
- Build rate-in minus rate-out models.
- Interpret time constants and lifetimes.
- Find equilibria of autonomous equations and classify their stability.
- Use the logistic model with harvesting to find maximum sustainable yield.
Key concepts
- Rate of change = rate in − rate out. Write the units of every term.
- A well-mixed lake with retention time τ loses a pollutant at rate C/τ once the inflow is clean.
- For an autonomous equation y′ = f(y), equilibria solve f(y) = 0; an equilibrium is stable if f′(y) < 0.
- Logistic growth P′ = rP(1 − P/K) levels off at K. With constant harvest H, the largest sustainable harvest is rK/4, at P = K/2.
Key equations
Worked examples, step by step
Example 2.1 Lake Erie’s retention time is 2.6 years and Lake Superior’s 191 years (EPA). If a pollutant’s inflow stops, how long until each lake’s concentration falls to 10%, treating each as well mixed?
Show the steps
- C/C₀ = e^(−t/τ) = 0.1, so t = τ ln 10.
- Erie: 2.6 × 2.303 = 5.99 years.
- Superior: 191 × 2.303 = 440 years.
Answer: About 6 years for Erie but about 440 for Superior. Prevention matters most where flushing is slow.
Example 2.2 EPA says methane lasts about a decade in the atmosphere on average. Treat a methane pulse as decaying with τ = 10 years. What fraction is left after 30 years?
Show the steps
- Fraction = e^(−30/10) = e^(−3).
Answer: About 5%. Cutting methane leaks therefore pays off within decades, unlike CO₂, which lingers far longer.
Example 2.3 A fish stock follows P′ = 0.4P(1 − P/10,000) − H (tonnes, years; sample values). Find the maximum sustainable yield.
Show the steps
- Growth rP(1 − P/K) is largest at P = K/2 = 5,000 t.
- There it equals rK/4 = 0.4 × 10,000/4.
Answer: 1,000 tonnes per year. Harvesting more than that drives the stock to collapse.
Practice problems
Work each on paper first, then check.
Coffee at 90 °C cools in a 20 °C room with T′ = −0.1(T − 20) (minutes). Find T(10).
Show answer
T = 20 + 70e^(−1) ≈ 45.8 °C.
A 100 L tank of fresh water gets brine of 0.5 g/L at 2 L/min and drains at 2 L/min. Find the salt after 50 minutes.
Show answer
x′ = 1 − x/50, x = 50(1 − e^(−t/50)); x(50) = 50(1 − e^(−1)) ≈ 31.6 g.
Find and classify the equilibria of y′ = y(1 − y).
Show answer
y = 0 (f′ = 1, unstable) and y = 1 (f′ = −1, stable).
In the fishery example, what happens if H = 1,200 t/yr?
Show answer
rP(1 − P/K) − H < 0 for every P, so the stock declines to zero.
Common misconceptions
Myth: Retention time is how long it takes to remove all the pollutant.
Actually: After one τ, 37% is still there; exponential decay never quite reaches zero.
Myth: More fishing effort always means more catch.
Actually: Past the maximum sustainable yield, more harvest shrinks the stock and future catch.
Myth: An equilibrium is always where the system ends up.
Actually: Only stable equilibria attract nearby solutions.
Flashcards
Balance law
Rate in − rate out.
Time constant
τ in e^(−t/τ); 63% of the change is done after τ.
Retention time
Lake volume ÷ outflow.
Equilibrium
f(y*) = 0.
Stability test
f′(y*) < 0 stable.
Max sustainable yield
rK/4 at P = K/2.
Self-check quiz
How this helps save the planet
The EPA lists Great Lakes retention times from 2.6 years (Erie) to 191 years (Superior), which is why toxins entering Lake Superior are a multi-generation problem.
Fishery scientists at NOAA assess stocks with population models built on these ideas; the logistic model’s maximum sustainable yield is the classic starting point.
On Saving Planets: Oceans hub · Rivers hub
Sources: U.S. EPA: Physical features of the Great Lakes (retention times) · U.S. EPA: Understanding global warming potentials · NOAA Fisheries: population assessments
Free learning resources
- Lebl, 1.6 Autonomous equations: Equilibria, logistic growth, harvesting.
- Lebl, 1.4 Integrating factor: Mixing problems.
- OpenStax Calculus 2, 4.4 The Logistic Equation: Logistic model.
Unit 3 · Weeks 4–5
Second-Order Linear Equations and Oscillations
Linear equations with constant coefficients, the characteristic equation, and free vibrations: undamped, underdamped, critically damped and overdamped.
Learning objectives
By the end of this unit you can:
- Solve ay″ + by′ + cy = 0 with the characteristic equation.
- Handle real, repeated and complex roots.
- Model mass-spring-damper systems.
- Classify damping and compute natural frequency and critical damping.
Key concepts
- Try y = e^(rt): ar² + br + c = 0. Two real roots give two exponentials; a repeated root gives (c₁ + c₂t)e^(rt); complex roots α ± iβ give e^(αt)(c₁ cos βt + c₂ sin βt).
- Superposition: sums of solutions of a homogeneous linear equation are solutions.
- mx″ + cx′ + kx = 0 has natural frequency ω₀ = √(k/m).
- Damping c < 2√(mk) oscillates and decays (underdamped); c = 2√(mk) returns fastest without overshoot (critical).
Key equations
Worked examples, step by step
Example 3.1 Solve y″ + 4y = 0, y(0) = 1, y′(0) = 0. What is the period?
Show the steps
- r² + 4 = 0, r = ±2i.
- y = cos 2t.
- Period 2π/2.
Answer: y = cos 2t, period π ≈ 3.14.
Example 3.2 Solve y″ + 2y′ + 5y = 0, y(0) = 1, y′(0) = 0, and find y(1).
Show the steps
- r = −1 ± 2i.
- y = e^(−t)(cos 2t + ½ sin 2t) fits both conditions.
- y(1) = e^(−1)(cos 2 + ½ sin 2).
Answer: ≈ 0.0142: the oscillation has nearly died out.
Example 3.3 A 2 kg mass hangs on a spring with k = 50 N/m. Find ω₀, the frequency in hertz, and the critical damping.
Show the steps
- ω₀ = √(50/2) = 5 rad/s.
- f = 5/(2π) = 0.796 Hz.
- c_crit = 2√(2 × 50) = 20 N·s/m.
Answer: 5 rad/s, 0.80 Hz, 20 N·s/m.
Practice problems
Work each on paper first, then check.
Solve y″ − 3y′ + 2y = 0.
Show answer
r = 1, 2: y = c₁eᵗ + c₂e^(2t).
Solve y″ + 2y′ + y = 0.
Show answer
Repeated r = −1: y = (c₁ + c₂t)e^(−t).
Classify x″ + 6x′ + 5x = 0.
Show answer
r = −1, −5: overdamped (c = 6 > 2√5).
Find the period of x″ + 9x = 0.
Show answer
ω = 3, period 2π/3 ≈ 2.09.
Common misconceptions
Myth: Complex roots mean complex solutions.
Actually: Real and imaginary parts give real solutions: e^(αt)cos βt and e^(αt)sin βt.
Myth: A repeated root gives only one solution.
Actually: The second is t e^(rt).
Myth: More damping always means faster settling.
Actually: Overdamped systems creep back slower than critically damped ones.
Flashcards
Characteristic equation
ar² + br + c = 0.
Superposition
Sums of solutions are solutions (homogeneous linear).
Repeated root
(c₁ + c₂t)e^(rt).
Natural frequency
ω₀ = √(k/m).
Underdamped
Oscillates while decaying.
Critical damping
c = 2√(mk).
Self-check quiz
How this helps save the planet
Every tall structure, from a wind-turbine tower to a bridge, has natural frequencies set by its stiffness and mass, exactly ω₀ = √(k/m) in the simplest model. Damping decides how fast gust-driven sway dies out, which affects fatigue and how long a structure lasts.
On Saving Planets: Physics I, Unit 8: oscillations
Sources: U.S. DOE: How do wind turbines work? · Lebl, 2.4 Mechanical vibrations
Free learning resources
- Lebl, 2.1 Second-order linear ODEs: Theory.
- Lebl, 2.2 Constant coefficient equations: Characteristic equation.
- Lebl, 2.4 Mechanical vibrations: Damping.
- PhET: Masses and Springs: Simulation.
Unit 4 · Week 6
Forced Vibration and Resonance
Nonhomogeneous equations, undetermined coefficients, the amplitude response of a driven oscillator, beats and resonance.
Learning objectives
By the end of this unit you can:
- Find particular solutions by undetermined coefficients.
- Write the general solution as homogeneous plus particular.
- Compute the steady-state amplitude of a driven damped oscillator.
- Explain resonance and how engineers avoid it.
Key concepts
- General solution = homogeneous solution + one particular solution.
- Guess the form of the forcing: for cos ωt try A cos ωt + B sin ωt.
- Driven at frequency ω, a damped oscillator settles into steady motion with amplitude X = F₀/√((k − mω²)² + (cω)²).
- Near ω = ω₀ the amplitude peaks; with damping ratio ζ = c/c_crit, the peak dynamic amplification is about 1/(2ζ). With no damping, the response grows without bound.
Key equations
Worked examples, step by step
Example 4.1 Find a particular solution of y″ + y = cos 2t.
Show the steps
- Try y_p = A cos 2t: −4A + A = 1.
- A = −1/3.
Answer: y_p = −⅓ cos 2t (out of phase, since the drive is above the natural frequency).
Example 4.2 Model a wind-turbine tower as a single oscillator with natural frequency 0.30 Hz and damping ratio ζ = 0.01 (sample values). The rotor turns at 12 rpm = 0.20 Hz, and three blades pass the tower at 0.60 Hz. Find the dynamic amplification at each, and at resonance.
Show the steps
- At 0.20 Hz, r = 0.667: 1/√((1 − 0.444)² + (0.0133)²) = 1.80.
- At 0.60 Hz, r = 2: 1/√(9 + 0.0016) = 0.333.
- At r = 1: 1/(2ζ) = 50.
Answer: 1.8 and 0.33, but 50 at resonance. That is why designers keep a tower’s natural frequency away from the rotor and blade-passing frequencies.
Example 4.3 Solve y″ + y = cos t, y(0) = y′(0) = 0.
Show the steps
- The forcing matches the natural frequency, so try y_p = t(A cos t + B sin t).
- Substituting gives A = 0, B = ½.
- y = ½ t sin t.
Answer: The amplitude grows like t/2 without limit: pure resonance.
Practice problems
Work each on paper first, then check.
Find y_p for y″ + 3y′ + 2y = 4.
Show answer
y_p = 2.
Find y_p for y″ − y = eᵗ.
Show answer
eᵗ solves the homogeneous equation, so y_p = ½ t eᵗ.
A 1 kg mass, k = 100 N/m, c = 2 N·s/m is driven by 10 cos 10t N. Find the steady amplitude.
Show answer
k − mω² = 0, so X = 10/(2 × 10) = 0.5 m.
For ζ = 0.05, what is the peak amplification?
Show answer
About 1/(2 × 0.05) = 10.
Common misconceptions
Myth: Resonance needs a large force.
Actually: A small periodic force at the natural frequency builds up a large response.
Myth: Damping removes resonance.
Actually: It caps the peak at about 1/(2ζ), which is still large for lightly damped structures.
Myth: Undetermined coefficients works for any forcing.
Actually: Only for polynomials, exponentials, sines and cosines and their products; otherwise use variation of parameters or Laplace.
Flashcards
Particular solution
Any one solution of the forced equation.
Undetermined coefficients
Guess the forcing’s form.
Amplitude response
X = F₀/√((k − mω²)² + (cω)²).
Damping ratio
ζ = c/(2√(mk)).
Resonance
Large response near ω₀; peak ≈ 1/(2ζ).
Beats
Two close frequencies make a slow envelope.
Self-check quiz
How this helps save the planet
Wind turbines turn ever larger and taller to harvest more energy, and their towers and blades are lightly damped. Avoiding resonance between the structure and the rotating rotor is a core part of keeping clean-energy machines safe and long-lived. Lab D explores it with a sample model.
On Saving Planets: Physics I, Unit 3: wind power
Sources: U.S. DOE: How do wind turbines work? · Lebl, 2.6 Forced oscillations and resonance
Free learning resources
- Lebl, 2.5 Nonhomogeneous equations: Undetermined coefficients and variation of parameters.
- Lebl, 2.6 Forced oscillations and resonance: Resonance.
- MIT OCW 18.03SC, Unit II: Second-order equations, lectures and problems.
Unit 5 · Weeks 7–8
Laplace Transforms
Transforming differential equations into algebra: the transform and its table, derivatives, step functions for switched inputs, impulses, and convolution.
Learning objectives
By the end of this unit you can:
- Compute Laplace transforms from the definition and a table.
- Solve initial value problems with transforms and partial fractions.
- Model switched inputs with unit step functions and shifts.
- Model sudden impulses with the delta function.
Key concepts
- L{f}(s) = ∫₀^∞ e^(−st) f(t) dt turns derivatives into multiplication by s, with initial values built in.
- Solve for Y(s), split with partial fractions, then invert with the table.
- A switch flipped at time a is u(t − a); the shift rule is L{u(t − a)f(t − a)} = e^(−as)F(s).
- An instant kick is δ(t − a), with L{δ(t − a)} = e^(−as).
Key equations
Worked examples, step by step
Example 5.1 Solve y′ − 2y = e^(3t), y(0) = 1.
Show the steps
- sY − 1 − 2Y = 1/(s − 3).
- Y = 1/((s − 2)(s − 3)) + 1/(s − 2) = [1/(s − 3) − 1/(s − 2)] + 1/(s − 2).
- Y = 1/(s − 3).
Answer: y = e^(3t). Check: 3e^(3t) − 2e^(3t) = e^(3t).
Example 5.2 A clean lake starts receiving a steady pollutant inflow at t = 2: y′ + y = u(t − 2), y(0) = 0. Find y(3).
Show the steps
- (s + 1)Y = e^(−2s)/s.
- 1/(s(s + 1)) = 1/s − 1/(s + 1), so y = (1 − e^(−(t − 2)))u(t − 2).
- y(3) = 1 − e^(−1).
Answer: ≈ 0.632 of the final level, one time constant after the switch.
Example 5.3 A resting oscillator y″ + 4y = δ(t − 1) gets a kick at t = 1. Find y(1 + π/4).
Show the steps
- (s² + 4)Y = e^(−s).
- y = ½ sin(2(t − 1)) u(t − 1).
- At t = 1 + π/4: ½ sin(π/2).
Answer: 0.5, the first peak.
Practice problems
Work each on paper first, then check.
Find L{t}.
Show answer
1/s².
Invert 3/(s² + 9).
Show answer
sin 3t.
Solve y′ + y = 0, y(0) = 4.
Show answer
Y = 4/(s + 1), y = 4e^(−t).
Solve y″ + y = 0, y(0) = 0, y′(0) = 2.
Show answer
Y = 2/(s² + 1), y = 2 sin t.
Find L{u(t − 3)}.
Show answer
e^(−3s)/s.
Common misconceptions
Myth: L{fg} = L{f}L{g}.
Actually: The product of transforms corresponds to convolution, not the product.
Myth: Initial conditions are applied at the end.
Actually: They enter immediately through L{y′} = sY − y(0).
Myth: Step functions need piecewise solving.
Actually: Laplace handles switches in one pass.
Flashcards
Laplace transform
∫₀^∞ e^(−st) f dt.
L{y′}
sY − y(0).
L{eᵃᵗ}
1/(s − a).
Unit step
u(t − a): off then on.
Delta
δ(t − a): instant impulse; L = e^(−as).
Convolution
L{f * g} = FG.
Self-check quiz
How this helps save the planet
Real environmental inputs switch on and off: a factory discharge starts, a dam release stops, a volcanic eruption injects aerosols in a single burst. Step and impulse inputs with Laplace transforms give the response to each in one calculation, the method engineers use for control systems in power grids and treatment plants.
On Saving Planets: Rivers hub
Sources: Lebl, 6.1 The Laplace transform · Lebl, 6.4 Dirac delta and impulse response
Free learning resources
- Lebl, 6.1 The Laplace transform: Definition and table.
- Lebl, 6.2 Transforms of derivatives and ODEs: Solving IVPs.
- Lebl, 6.3 Convolution: Convolution.
- Lebl, 6.4 Dirac delta: Impulses.
Unit 6 · Weeks 9–10
Linear Systems, Eigenvalues and Phase Planes
Systems x′ = Ax, eigenvalue methods, the phase plane, and the classification of equilibria as nodes, saddles, spirals and centers.
Learning objectives
By the end of this unit you can:
- Write coupled equations and higher-order equations as first-order systems.
- Solve x′ = Ax with eigenvalues and eigenvectors.
- Sketch phase portraits.
- Classify the origin with the trace and determinant.
Key concepts
- If Av = λv, then x = e^(λt)v solves x′ = Ax; the general solution combines one such term per eigenvalue.
- Real eigenvalues of opposite sign: a saddle. Both negative: a stable node.
- Complex eigenvalues α ± iβ: spirals (stable if α < 0); centers if α = 0.
- For a 2 × 2 matrix, λ² − (tr A)λ + det A = 0.
Key equations
Worked examples, step by step
Example 6.1 A two-box carbon model: atmosphere A and ocean O exchange carbon with A′ = −0.2A + 0.1O, O′ = 0.2A − 0.1O (per year; sample rates). Start with 3 units in the air and none in the ocean.
Show the steps
- Eigenvalues: tr = −0.3, det = 0, so λ = 0 and −0.3.
- λ = 0 has eigenvector (1, 2): the equilibrium split is 1 : 2. Total carbon (3) is conserved, so A → 1, O → 2.
- A(t) = 1 + 2e^(−0.3t).
Answer: The air keeps one third of the pulse in this toy model, with time scale 1/0.3 ≈ 3.33 years.
Example 6.2 Classify the origin for A = [[1, 2], [2, 1]].
Show the steps
- λ² − 2λ − 3 = 0, λ = 3, −1.
Answer: Opposite signs: a saddle (unstable).
Example 6.3 Classify the origin for A = [[−1, −2], [2, −1]].
Show the steps
- tr = −2, det = 5: λ² + 2λ + 5 = 0.
- λ = −1 ± 2i.
Answer: A stable spiral: solutions circle inward, decaying like e^(−t).
Practice problems
Work each on paper first, then check.
Write y″ + 3y′ + 2y = 0 as a system.
Show answer
x₁ = y, x₂ = y′: x₁′ = x₂, x₂′ = −2x₁ − 3x₂.
Eigenvalues of [[2, 0], [0, −3]]?
Show answer
2 and −3: a saddle.
Classify [[0, 1], [−4, 0]].
Show answer
λ = ±2i: a center (undamped oscillation).
Classify [[−3, 1], [0, −2]].
Show answer
λ = −3, −2: a stable node.
Common misconceptions
Myth: Eigenvectors are unique.
Actually: Any nonzero multiple works.
Myth: A zero eigenvalue means no motion.
Actually: It means a line of equilibria or a conserved quantity, as in the carbon box model.
Myth: Complex eigenvalues give complex solutions only.
Actually: Real and imaginary parts give real solutions.
Flashcards
Linear system
x′ = Ax.
Eigenvalue method
x = e^(λt)v.
Saddle
Real eigenvalues of opposite sign.
Node
Real eigenvalues of the same sign.
Spiral
Complex eigenvalues, α ≠ 0.
Trace–determinant
λ² − tr λ + det = 0.
Self-check quiz
How this helps save the planet
Carbon-cycle models are linear systems of boxes (atmosphere, surface ocean, deep ocean, land) exchanging carbon. Their eigenvalues are the time scales on which a CO₂ pulse is absorbed: some fast, some taking centuries or longer. NASA’s carbon-cycle overview describes the fast and slow cycles these boxes represent.
On Saving Planets: Oceans hub
Sources: NASA Earth Observatory: The Carbon Cycle · NOAA Global Monitoring Laboratory: Trends in atmospheric CO₂
Free learning resources
- Lebl, 3.1 Introduction to systems: Systems.
- Lebl, 3.4 Eigenvalue method: Eigenvalues.
- Lebl, 3.5 Two-dimensional systems and their vector fields: Phase portraits.
Unit 7 · Weeks 11–12
Nonlinear Systems and Stability
Equilibria of nonlinear systems, linearization with the Jacobian, predator-prey cycles, and the SIR epidemic model.
Learning objectives
By the end of this unit you can:
- Find equilibria of nonlinear systems.
- Linearize with the Jacobian and classify equilibria.
- Analyze the Lotka–Volterra predator-prey model.
- Use the SIR model: R₀, herd-immunity threshold and final size.
Key concepts
- Near an equilibrium, a nonlinear system behaves like its linearization x′ = J x, where J is the Jacobian (unless the linearization is a center).
- Lotka–Volterra: prey x′ = ax − bxy, predators y′ = −cy + dxy. The equilibrium is (c/d, a/b); populations cycle around it.
- SIR: S′ = −βSI, I′ = βSI − γI, R′ = γI. The basic reproduction number R₀ = β/γ; an outbreak grows only if R₀S > 1.
- The herd-immunity threshold is 1 − 1/R₀; the final uninfected share s∞ solves ln s∞ = R₀(s∞ − 1).
Key equations
Worked examples, step by step
Example 7.1 Prey and predators follow x′ = x − 0.02xy, y′ = −0.5y + 0.005xy (sample values, years). Find the coexistence equilibrium and the period of small cycles.
Show the steps
- x′ = 0 needs y = 1/0.02 = 50; y′ = 0 needs x = 0.5/0.005 = 100.
- The Jacobian there has eigenvalues ±i√(ac) = ±i√0.5.
- Period 2π/√0.5.
Answer: 100 prey and 50 predators; small cycles last about 8.9 years, with predators lagging prey.
Example 7.2 An epidemic has β = 0.3 per day and γ = 0.1 per day. Find R₀, the herd-immunity threshold, and the final share ever infected.
Show the steps
- R₀ = 3.
- Threshold 1 − 1/3 = 0.667.
- Iterate s = e^(3(s − 1)): s∞ = 0.0595.
Answer: R₀ = 3; immunity in 67% stops growth; with no intervention about 94% are eventually infected, more than the threshold because the epidemic overshoots.
Example 7.3 Classify the equilibrium P = K of logistic growth P′ = rP(1 − P/K).
Show the steps
- f′(P) = r − 2rP/K; f′(K) = −r < 0.
Answer: Stable: small disturbances decay at rate r.
Practice problems
Work each on paper first, then check.
Find the equilibria of x′ = x(3 − x − y), y′ = y(2 − x − y).
Show answer
(0, 0), (3, 0), (0, 2) (no interior one: the lines x + y = 3 and x + y = 2 are parallel).
Linearize x′ = sin x near x = 0 and x = π.
Show answer
f′(0) = 1 (unstable), f′(π) = −1 (stable).
If R₀ = 2, what is the herd-immunity threshold?
Show answer
1 − 1/2 = 50%.
In the predator-prey example, what happens to the prey if predators are removed?
Show answer
x′ = x: exponential growth (until other limits apply).
Common misconceptions
Myth: Linearization always predicts the behavior.
Actually: Not when it gives a center; the nonlinear terms decide.
Myth: Herd immunity means the epidemic stops at once.
Actually: Infections decline after the threshold but can still overshoot.
Myth: Predator-prey cycles mean the model is unstable.
Actually: Lotka–Volterra orbits are closed: neutrally stable, neither growing nor shrinking.
Flashcards
Jacobian
Matrix of first partials.
Linearization
x′ ≈ J x near an equilibrium.
Lotka–Volterra
Predator-prey cycles around (c/d, a/b).
SIR
Susceptible, infected, recovered.
R₀
β/γ, new infections per case.
Herd-immunity threshold
1 − 1/R₀.
Self-check quiz
How this helps save the planet
On Isle Royale in Lake Superior, the National Park Service charts wolf and moose numbers back to 1980, a long predator-prey record. Their linked ups and downs, and the Park Service’s decision to relocate new wolves to the island, are exactly the kind of system Lotka–Volterra models begin to explain.
The same SIR mathematics that guides public-health response is used for wildlife diseases that threaten ecosystems.
On Saving Planets: Oceans hub
Sources: National Park Service: Wolves of Isle Royale · Wolves and Moose of Isle Royale project (long-term data)
Free learning resources
- Lebl, 8.1 Linearization, critical points and equilibria: Linearization.
- Lebl, 8.2 Stability and classification: Stability.
- Lebl, 8.3 Applications of nonlinear systems: Predator-prey and more.
Unit 8 · Week 13
Numerical Methods
Euler, improved Euler and Runge–Kutta methods, step size and error, and a climate energy-balance model solved numerically.
Learning objectives
By the end of this unit you can:
- Apply Euler, improved Euler (Heun) and RK4.
- Explain the order of a method and how error scales with step size.
- Solve systems numerically.
- Build and solve a zero-dimensional energy-balance climate model.
Key concepts
- Euler steps along the slope: y_{n+1} = y_n + h f(t_n, y_n). Its global error is proportional to h.
- Improved Euler averages two slopes (error ∝ h²); RK4 averages four (error ∝ h⁴).
- Halving h cuts RK4 error about 16 times, which is why it is the workhorse.
- Energy balance: C dT/dt = absorbed sunlight − εσT⁴. Its equilibrium and time scale come from the equation directly.
Key equations
Worked examples, step by step
Example 8.1 Approximate y(1) for y′ = y, y(0) = 1 with h = 0.1 using Euler, improved Euler and RK4.
Show the steps
- Each Euler step multiplies by 1.1: 1.1¹⁰ = 2.5937.
- Improved Euler multiplies by 1.105: 2.7141.
- RK4 multiplies by 1.1051708: 2.7182797.
Answer: True e = 2.7182818. Errors: 0.12, 0.0042 and 0.000002.
Example 8.2 Earth absorbs 340 × (1 − 0.29) = 241.4 W/m² of sunlight (NASA). If it radiated as a blackbody (ε = 1), what temperature would balance it? What effective ε gives today’s average of about 288 K?
Show the steps
- σT⁴ = 241.4 gives T = (241.4/5.67 × 10⁻⁸)^(1/4) = 255 K.
- ε = 241.4/(σ × 288⁴) = 0.619.
Answer: 255 K (−18 °C) without a greenhouse effect; an effective emissivity of 0.62 represents the greenhouse effect that keeps Earth habitable.
Example 8.3 With ε = 0.619 at 288 K and an ocean mixed layer 50 m deep (C = 1,000 × 4,186 × 50 J/(m²·K); sample depth), find λ, the time scale, and the no-feedback warming for a 4 W/m² forcing (sample value).
Show the steps
- λ = 4εσT³ = 3.35 W/(m²·K).
- τ = C/λ = 6.24 × 10⁷ s ≈ 1.98 years (about 2).
- ΔT = 4/3.35.
Answer: About 1.2 K of warming, approached on a 2-year time scale in this simple model. Real feedbacks (water vapor, ice) make the warming larger, as the IPCC assesses.
Practice problems
Work each on paper first, then check.
One Euler step for y′ = t + y, y(0) = 1, h = 0.1.
Show answer
y₁ = 1 + 0.1(0 + 1) = 1.1.
If Euler’s error is 0.12 at h = 0.1, estimate it at h = 0.05.
Show answer
About 0.06 (first order).
If RK4’s error is 2 × 10⁻⁶ at h = 0.1, estimate it at h = 0.05.
Show answer
About 2 × 10⁻⁶/16 ≈ 1.3 × 10⁻⁷.
What equilibrium temperature does the EBM give with ε = 1 and albedo 0.29?
Show answer
About 255 K.
Common misconceptions
Myth: Smaller steps are always better.
Actually: Round-off and cost grow; a higher-order method is usually better than tiny Euler steps.
Myth: A numerical answer is exact if the code runs.
Actually: Always check convergence by halving h.
Myth: The 255 K result means the model is wrong.
Actually: It shows how much the greenhouse effect warms Earth: about 33 K.
Flashcards
Euler’s method
y + h f.
Improved Euler
Average of two slopes; order 2.
RK4
Four slopes; order 4.
Step size
h; check convergence by halving.
Energy-balance model
C T′ = S(1 − α) − εσT⁴.
Climate time scale
τ = C/λ.
Self-check quiz
How this helps save the planet
Every climate model, from this one-line energy balance to the global models assessed by the IPCC, is a system of differential equations solved numerically. NASA’s energy budget gives the inputs: 340 W/m² of sunlight arrives on average and 29% is reflected.
On Saving Planets: Intro to Programming, simulation unit
Sources: NASA Earth Observatory: Climate and Earth’s Energy Budget · IPCC AR6 Working Group I: The Physical Science Basis
Free learning resources
- Lebl, 1.7 Numerical methods: Euler’s method: Euler and error.
- Trench, Ch. 3: Numerical methods: Improved Euler and RK4.
Unit 9 · Week 14
Series Solutions, Fourier Series and the Heat Equation
Power-series solutions of ODEs, Fourier series of periodic functions, and separation of variables for the heat equation.
Learning objectives
By the end of this unit you can:
- Find power-series solutions from a recurrence.
- Compute Fourier coefficients of simple periodic functions.
- Solve the heat equation on a rod by separation of variables.
- Estimate how deep seasonal temperature waves reach into the ground.
Key concepts
- Substitute y = Σaₙxⁿ, match powers, and get a recurrence for the coefficients.
- A periodic function is a sum of sines and cosines; for an odd function on (−π, π), bₙ = (2/π)∫₀^π f sin nx dx.
- The heat equation u_t = α u_xx spreads heat; each sine mode decays like e^(−α(nπ/L)²t), fast modes first.
- A periodic surface temperature penetrates the ground with damping depth d = √(2α/ω).
Key equations
Worked examples, step by step
Example 9.1 Solve y″ + y = 0 by power series.
Show the steps
- Σ(n + 2)(n + 1)aₙ₊₂xⁿ + Σaₙxⁿ = 0.
- aₙ₊₂ = −aₙ/((n + 2)(n + 1)).
- Even terms build cos x, odd terms sin x.
Answer: y = a₀ cos x + a₁ sin x, recovered from the recurrence.
Example 9.2 Find the Fourier series of the square wave f = 1 on (0, π), −1 on (−π, 0).
Show the steps
- Odd, so only sines: bₙ = (2/π)∫₀^π sin nx dx = (2/(nπ))(1 − cos nπ).
- bₙ = 4/(nπ) for odd n, 0 for even n.
Answer: f = (4/π)(sin x + sin 3x/3 + sin 5x/5 + …); b₁ ≈ 1.273.
Example 9.3 Soil with diffusivity α = 5 × 10⁻⁷ m²/s (sample value) feels a yearly surface temperature cycle. Find the damping depth.
Show the steps
- ω = 2π/(3.156 × 10⁷ s) = 1.99 × 10⁻⁷ s⁻¹.
- d = √(2 × 5 × 10⁻⁷/1.99 × 10⁻⁷).
Answer: About 2.24 m: the yearly swing shrinks by a factor e for each 2.2 m of depth, so deeper ground responds to warming slowly, over many years.
Practice problems
Work each on paper first, then check.
Find the recurrence for y′ = y by power series.
Show answer
aₙ₊₁ = aₙ/(n + 1), so y = a₀eˣ.
Is f(x) = x² odd or even? Which Fourier terms appear?
Show answer
Even: cosines only (plus a constant).
A rod mode decays like e^(−α(π/L)²t). How does doubling L change its decay time?
Show answer
It multiplies the decay time by 4.
Find b₃ for the square wave.
Show answer
4/(3π) ≈ 0.424.
Common misconceptions
Myth: Fourier series only work for smooth functions.
Actually: They work for jumps too, converging to the midpoint at a jump.
Myth: Heat diffuses at a fixed speed.
Actually: Diffusion time grows with distance squared.
Myth: Series solutions are always better than formulas.
Actually: They are most useful when no closed form exists, as for Airy or Bessel equations.
Flashcards
Power-series method
Substitute Σaₙxⁿ; get a recurrence.
Fourier series
Sum of sines and cosines.
Fourier coefficient
bₙ = (2/π)∫₀^π f sin nx dx.
Heat equation
u_t = α u_xx.
Separation of variables
u = X(x)T(t).
Damping depth
√(2α/ω).
Self-check quiz
How this helps save the planet
Permafrost in the Northern Hemisphere holds about 1,672 billion tons of organic carbon, according to NASA. How fast surface warming reaches it is a heat-equation problem, and the damping depth shows why deeper layers respond slowly but keep responding long after the surface warms.
On Saving Planets: Calculus III
Sources: NASA Earth Observatory: The Carbon Cycle · Lebl, 4.6 PDEs, separation of variables, and the heat equation
Free learning resources
- Lebl, 7.1 Power series: Power series.
- Lebl, 7.2 Series solutions of linear second-order ODEs: Series solutions.
- Lebl, 4.2 Trigonometric series: Fourier series.
- Lebl, 4.6 The heat equation: Heat equation.
Computer labs
Four computer labs: an SIR epidemic solved with Euler and RK4, a predator-prey phase plane, a climate energy-balance model, and a frequency sweep of a wind-turbine tower model.
Sample parameters are labeled as such; each lab links to a public source for the real system.
Safety reference: Lebl: Notes on Diffy Qs (free textbook)
Lab A: An SIR epidemic with Euler and RK4
Week 12
Goal: Solve the SIR model numerically, find the peak, and compare Euler with RK4 against the exact final size.
Procedure
- Start with S = 0.999, I = 0.001, R = 0 (fractions of the population).
- Run Euler with h = 1 day and h = 0.1 day to day 160, then RK4 with h = 1 day.
- Record the final R and the peak of I and when it occurs; compare with theory.
Data table
| Method | Final R (day 160) |
|---|---|
| Euler, h = 1 | 0.9467 |
| Euler, h = 0.1 | 0.9411 |
| RK4, h = 1 | 0.9405 |
| Exact final size (theory) | 0.9405 |
Computed values for sample rates; not a real outbreak.
Sample calculation
- The peak is 30.1% infected at once, near day 38.4; theory gives I_max = 1 − (1 + ln R₀)/R₀ = 0.300.
- Euler with h = 1 overstates the final size by 0.006; RK4 with the same step matches theory to 4 digits.
Safety
- None beyond normal computer use.
Home or virtual alternative
- A spreadsheet works; Python is easier for RK4.
Lebl, 8.3 Applications of nonlinear systems
Planet link: The same model structure is used for disease outbreaks in people and in wildlife.
Lab B: The predator-prey phase plane
Week 11
Goal: Integrate the Lotka–Volterra model with RK4 and plot the closed orbit in the phase plane.
Procedure
- Start at 150 prey and 50 predators.
- Use RK4 with a small step (0.001 years in Python, or 0.01 in a spreadsheet) for 40 years.
- Plot y against x and record the prey range and the period.
Data table
| Quantity | Value |
|---|---|
| Equilibrium | (100, 50) |
| Prey range | 62.6 to 150 |
| Period | 8.99 years |
| Small-cycle period (linear theory) | 8.89 years |
Computed values; sample parameters.
Sample calculation
- The orbit closes: the populations repeat every 8.99 years, slightly longer than the 8.89 years linear theory predicts for small cycles, because this cycle is large.
- Predators peak after prey, a lag seen in many real predator-prey records.
Safety
- None beyond normal computer use.
Home or virtual alternative
- Desmos can draw the phase plane slope field.
Wolves and Moose of Isle Royale project (long-term data)
Planet link: Long-term records such as Isle Royale’s wolves and moose test these models. National Park Service: Wolves of Isle Royale
Lab C: A climate energy-balance model
Week 13
Goal: Step a one-box climate model through time after a sudden forcing and find its time scale.
Procedure
- Use C = 2.09 × 10⁸ J/(m²·K) (50 m of ocean), λ = 3.35 W/(m²·K) from Unit 8, and a sudden forcing F = 4 W/m² (sample value).
- Step with Euler using monthly steps for 10 years.
- Compare with the exact solution ΔT = (F/λ)(1 − e^(−t/τ)).
Data table
| Year | ΔT (K) |
|---|---|
| 1 | 0.48 |
| 2 | 0.77 |
| 5 | 1.10 |
| 10 | 1.19 |
Computed model values; no feedbacks other than radiation.
Sample calculation
- Equilibrium ΔT = F/λ = 1.19 K; τ = C/λ ≈ 2.0 years, so the warming is about 63% done after 2 years in this model.
- Real climate responds more slowly and warms more, because heat mixes into the deep ocean and feedbacks amplify it.
Safety
- None beyond normal computer use.
Home or virtual alternative
- A spreadsheet is enough.
NASA Earth Observatory: Climate and Earth’s Energy Budget
Planet link: Energy-balance models are the first step toward the climate projections assessed by the IPCC. IPCC AR6 Working Group I: The Physical Science Basis
Lab D: Resonance sweep of a wind-turbine tower model
Week 6
Goal: Compute the dynamic amplification of a single-oscillator tower model across driving frequencies.
Procedure
- Use f₀ = 0.30 Hz and ζ = 0.01 (sample values).
- Tabulate A for driving frequencies from 0.1 to 0.6 Hz.
- Mark the rotor frequency (0.20 Hz at 12 rpm) and the blade-passing frequency (0.60 Hz) and judge the design.
Data table
| Driving frequency (Hz) | Amplification |
|---|---|
| 0.10 | 1.12 |
| 0.20 | 1.80 |
| 0.25 | 3.27 |
| 0.30 | 50.0 |
| 0.35 | 2.76 |
| 0.60 | 0.333 |
Computed values for a sample model, not a specific turbine.
Sample calculation
- Both operating frequencies sit well away from the 0.30 Hz peak, where the response would be 50 times the static deflection.
- If the rotor sped up to 18 rpm (0.30 Hz), it would drive the tower at resonance; this is why designers check the natural frequency against the whole speed range.
Safety
- None beyond normal computer use.
Home or virtual alternative
- A spreadsheet is enough.
Planet link: Avoiding resonance keeps wind turbines generating clean power safely for decades. U.S. DOE: How do wind turbines work?
Exam prep practice sets
Time yourself: about 10 minutes per problem. Then compare with the full solutions.
Midterm practice set (Units 1–5)
Solve y′ = −2ty, y(0) = 3, and find y(1).
Show full solution
- dy/y = −2t dt; y = 3e^(−t²).
Answer: 3e^(−1) ≈ 1.104.
A lake with retention time 4 years: how long until a pollutant falls to half after inflow stops?
Show full solution
- t = 4 ln 2.
Answer: ≈ 2.77 years.
Solve y″ + 6y′ + 9y = 0, y(0) = 1, y′(0) = 0, and find y(1).
Show full solution
- y = (1 + 3t)e^(−3t).
Answer: 4e^(−3) ≈ 0.199.
Solve y′ + y = 1, y(0) = 0 by Laplace and find y(2).
Show full solution
- Y = 1/(s(s + 1)) = 1/s − 1/(s + 1).
Answer: 1 − e^(−2) ≈ 0.865.
Final practice set (all units)
Classify the origin of x′ = [[0, 1], [−2, −2]]x.
Show full solution
- λ² + 2λ + 2 = 0, λ = −1 ± i.
Answer: Stable spiral.
An SIR epidemic has β = 0.5, γ = 0.25 per day. Find R₀ and the herd-immunity threshold.
Show full solution
- R₀ = 2.
Answer: R₀ = 2, threshold 50%.
Two Euler steps with h = 0.5 for y′ = −y, y(0) = 1.
Show full solution
- Each step multiplies by 0.5.
Answer: y(1) ≈ 0.25 (true e^(−1) ≈ 0.368).
A mass-spring system with m = 1 kg, k = 16 N/m, c = 0.4 N·s/m. Find ω₀ and ζ.
Show full solution
- ω₀ = 4 rad/s; c_crit = 8.
Answer: ω₀ = 4 rad/s, ζ = 0.05 (peak amplification about 10).
Formula sheet
Constants
| Constant | Value |
|---|---|
| Stefan–Boltzmann constant σ | 5.67 × 10⁻⁸ W/(m²·K⁴) |
| Average sunlight at the top of the atmosphere | 340 W/m² |
| Fraction of sunlight reflected | 29% |
| Lake Superior retention time | 191 years |
| Lake Erie retention time | 2.6 years |
| Methane’s average lifetime | about a decade |
Source: NASA Earth Observatory: Climate and Earth’s Energy Budget
Free texts and tools
Use these free textbooks and tools alongside the course.
- Lebl: Notes on Diffy Qs: Free, complete textbook for this course.
- Trench: Elementary Differential Equations (free text): Free textbook with many exercises.
- MIT OpenCourseWare 18.03SC Differential Equations: Free MIT lectures, recitations and problem sets.
- Desmos graphing calculator: Graph solutions and slope fields.
- PhET: Masses and Springs: Damping and oscillation intuition.
Glossary
- Autonomous equation
- y′ = f(y), with no explicit t.
- Characteristic equation
- ar² + br + c = 0 for constant-coefficient ODEs.
- Critical damping
- Fastest return without overshoot.
- Eigenvalue
- λ with Av = λv.
- Energy-balance model
- Climate model balancing absorbed sunlight and emitted heat.
- Equilibrium
- Constant solution.
- Euler’s method
- First-order numerical stepping.
- Exact equation
- M dx + N dy = 0 with M_y = N_x.
- Fourier series
- Expansion in sines and cosines.
- Heat equation
- u_t = α u_xx.
- Integrating factor
- Multiplier making a linear ODE exact.
- Jacobian
- Matrix of first partials used to linearize.
- Laplace transform
- ∫₀^∞ e^(−st) f dt.
- Lotka–Volterra
- Predator-prey model.
- Phase portrait
- Trajectories in the state plane.
- Resonance
- Large response when forcing matches a natural frequency.
- Runge–Kutta (RK4)
- Fourth-order numerical method.
- R₀
- Basic reproduction number.
- SIR model
- Susceptible–infected–recovered epidemic model.
- Separable equation
- dy/dx = g(x)h(y).
Self-assessment rubric
No grade or credit is given. Use this to judge honestly where you are in each unit.
| Level | What it looks like |
|---|---|
| 4 · Mastery | Builds models from a description, solves them by formula and numerically, checks units and long-run behavior. |
| 3 · Proficient | Solves the practice problems with occasional slips. |
| 2 · Developing | Follows worked examples but struggles to set up models or choose a method. |
| 1 · Beginning | Applies formulas without checking solutions; needs to revisit the unit. |
Suggested weighting if you want a score
| Part | Weight |
|---|---|
| Practice problems | 30% |
| Computer labs | 15% |
| Midterm practice set | 25% |
| Final practice set | 30% |
Aim for level 3 or better in every unit before moving to the next course.