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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 set0/3Practice∞
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 set0/1Practice∞
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 set0/1Practice∞

Euler's Method and Its Three Equations

How does following one line element at a time turn a direction field into the three recursive equations a computer actually runs?


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    1. 1

      The initial value problem (IVP)

      Write the pair y′=f(x,y)y' = f(x,y)y′=f(x,y) and y(x0)=y0y(x_0) = y_0y(x0​)=y0​, name the two parts, and note that numerical methods are how such problems are actually solved in practice.

    2. 2

      Euler's method geometrically

      Describe starting at (x0,y0)(x_0, y_0)(x0​,y0​), following the line element of slope f(x,y)f(x,y)f(x,y) for one step of size hhh, recomputing the slope at the new point, and repeating to produce a broken-line approximation.

    3. 3

      Deriving the three Euler equations

      Complete the triangle at (xn,yn)(x_n, y_n)(xn​,yn​) with slope AnA_nAn​, horizontal side hhh and vertical side hAnh A_nhAn​, then write out xn+1=xn+hx_{n+1} = x_n + hxn+1​=xn​+h, yn+1=yn+hAny_{n+1} = y_n + h A_nyn+1​=yn​+hAn​, and An=f(xn,yn)A_n = f(x_n, y_n)An​=f(xn​,yn​).

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