Understanding Improved Euler's Method (Differential Equations 44)
Predictor-Corrector Averages Two Slopes
To fix the flat-slope problem, improved Euler's method uses a predictor-corrector model described as two steps forward and one step back. First, it uses standard Euler's method to predict a point at x_(n+1) and plugs that predicted point back into the differential equation to find a second, steeper slope. Averaging the shallow initial slope and the steep predicted slope produces a corrected trajectory much closer to the true curve.
Quadratic Accuracy Gains and U-Sub Notation
Professor Leonard explains that dividing the step size by 10 makes improved Euler's method 100 times more accurate, and cutting h in half gives four times the accuracy for twice the work. To avoid confusing the predicted y_(n+1) with the final corrected y_(n+1), the method renames the temporary Euler prediction u_(n+1), functioning like a temporary u-substitution in integration.
Deriving the Predictor-Corrector Formula
Leonard writes out the formal algorithm: first calculate the predictor u_(n+1) = y_n + h * f(x_n, y_n), then evaluate the second slope at the new coordinates f(x_(n+1), u_(n+1)). He warns students not to accidentally plug the old x_n into the second slope. Finally, the corrected value is y_(n+1) = y_n + (h/2) * [f(x_n, y_n) + f(x_(n+1), u_(n+1))], placing the 2 under h to simplify calculator work.
Computing the First Step for y' = 2y
To test the formula, Leonard sets up the initial value problem y' = 2y with y(0) = 1/2 and step size h = 0.1 to approximate y(0.5) over five iterations, noting the exact solution is y = (1/2)e^(2x). For the first step, the predictor gives u_1 = 1/2 + 0.1 * (2 * 1/2) = 0.6. Plugging u_1 = 0.6 into the corrector formula sets up y_1 = 1/2 + (0.1/2) * [2(1/2) + 2(0.6)].
Five-Step Result Beats Euler by Over 10x
Completing all five iterations to x_5 = 0.5, improved Euler's method yields y_5 ≈ 1.35135 compared to the exact analytical value y(0.5) ≈ 1.35914, missing by only about 8 thousandths. On the same problem, five iterations of standard Euler's method gave 1.24416, off by about 11 hundredths. Leonard concludes that mastering Euler's method provides the essential foundation for improved Euler's and the upcoming Runge-Kutta method.







