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Differential Equations
The Geometric View of ODEs
01Direction Fields and the Geometric View of ODEs02Integral Curves and the Isocline Method03Perpendicular Isoclines: The Circles of y' = -x/y04Isocline Corridors and a Trap for Solutions05Existence and Uniqueness: One Curve Through a Point06Where Existence and Uniqueness FailProblem set0/10Problem set 20/10MIT problem set0/3
Numerical Methods
01Euler's Method and Its Three Equations02Euler's Method by Hand: Building the Table03Convexity and the Sign of Euler's Error04First Order Error: Halve the Step, Halve the Error05Improved Euler (RK2): Averaging Two Slopes06Runge-Kutta 4 and the Price of Accuracy07Singular Points: Where Every Method FailsProblem set0/10Problem set 20/10MIT problem set0/1
First-Order Linear Equations
01First Order Linear: Standard Form and the Sign of p02The Conduction-Diffusion Model: Newton's Law of Cooling03The Integrating Factor and the Four-Step Method04Integrating Factors: Two Mistakes and a Trick Question05Constant Coefficients: Steady State and TransientProblem set0/10Problem set 20/10MIT problem set0/1
Substitution Methods
01Scaling a First Order ODE: The Fourth-Power Cooling Law02The Bernoulli Equation and the Substitution v = y^(1-n)03A Worked Bernoulli Example: The Substitution v = 1/y04Homogeneous Equations: The Zoom Test and y = zx05Modelling the Lighthouse Chase: A 45-Degree Heading06The Boat's Path: A Logarithmic Spiral in Polar CoordinatesProblem set0/10Problem set 20/10
Autonomous Equations
01Autonomous Equations and Time-Translation Invariance02Critical Points: Constant Solutions as Barriers03The Embezzler Model dy/dt = ry - w: One Unstable Balance04The Logistic Equation: Carrying Capacity Without Solving05Stable, Unstable and Semi-Stable Critical Points06Maximum Sustainable Harvest: Why h = a^2/4bProblem set0/10Problem set 20/10MIT problem set0/3
Complex Numbers
01Complex Division, Polar Form and the Definition of e^(iθ)02Why e^(iθ) Earns Its Name: Multiply Moduli, Add Angles03Integrating e^(-x) cos x by Complexifying04The nth Roots of Unity: Zeta on the Unit CircleProblem set0/10Problem set 20/10MIT problem set0/4
Input and Response
01Steady-State and Transient Solutions of y' + ky = q(t)02Input, Response and the Superposition Principle03Solving y' + ky = k cos ωt by Complexification04Amplitude and Phase Lag of a Sinusoidal ResponseProblem set0/10Problem set 20/10MIT problem set0/2
01Cartesian Method: Rationalise, Then Take the Real Part02The Sinusoidal Identity and Its Right Triangle03Three Proofs of the Sinusoidal Identity04The Mixing Tank: Salt Balance and k = r/V05RC Circuits, Decay Chains and Negative kProblem set0/10Problem set 20/10MIT problem set0/2
Second-Order Linear Equations
01Second-Order Linear ODEs and the Spring–Mass–Dashpot02The Characteristic Equation and Overdamped Motion03Complex Roots a ± bi and Underdamped Oscillation04Critical Damping and Reduction of Order: t e^(−at)Problem set0/10Problem set 20/10MIT problem set0/2

Complex Roots a ± bi and Underdamped Oscillation

How do the real and imaginary parts of one complex exponential give two real solutions, and why do complex roots mean a decaying oscillation?


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Your summary note

    1. 1

      Real and imaginary parts of a complex solution are solutions

      State the theorem for y′′+Ay′+By=0y'' + Ay' + By = 0y′′+Ay′+By=0 with A,BA, BA,B real, write out the proof by splitting into real and imaginary parts, and mark the step that needs A,BA, BA,B real.

    2. 2

      General solution y=eat(c1cos⁡bt+c2sin⁡bt)y = e^{at}(c_1\cos bt + c_2\sin bt)y=eat(c1​cosbt+c2​sinbt)

      Expand e(a+bi)te^{(a+bi)t}e(a+bi)t by Euler's formula to get y1=eatcos⁡bty_1 = e^{at}\cos bty1​=eatcosbt and y2=eatsin⁡bty_2 = e^{at}\sin bty2​=eatsinbt, then rewrite c1cos⁡bt+c2sin⁡btc_1\cos bt + c_2\sin btc1​cosbt+c2​sinbt as Ccos⁡(bt−ϕ)C\cos(bt - \phi)Ccos(bt−ϕ), (C,ϕ)(C, \phi)(C,ϕ) being polar coordinates of (c1,c2)(c_1, c_2)(c1​,c2​).

    3. 3

      Worked example y′′+4y′+5y=0y'' + 4y' + 5y = 0y′′+4y′+5y=0

      Solve the characteristic equation to r=−2±ir = -2 \pm ir=−2±i, write the general solution, and apply y(0)=1y(0) = 1y(0)=1, y′(0)=0y'(0) = 0y′(0)=0 to reach y=e−2t(cos⁡t+2sin⁡t)y = e^{-2t}(\cos t + 2\sin t)y=e−2t(cost+2sint).

    4. 4

      The underdamped picture: oscillation inside a decaying envelope

      Convert the example to 5 e−2tcos⁡(t−ϕ)\sqrt{5}\,e^{-2t}\cos(t - \phi)5​e−2tcos(t−ϕ) with tan⁡ϕ=2\tan\phi = 2tanϕ=2, ϕ≈63∘\phi \approx 63^\circϕ≈63∘, and sketch it between the envelopes ±5 e−2t\pm\sqrt{5}\,e^{-2t}±5​e−2t, marking the first zero at π/2+ϕ\pi/2 + \phiπ/2+ϕ.

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