Modified Euler method (1st order derivative) Formula & Examples
atozmath.com › CONM › RungeKuttaHere, x0 = 0, y0 = 1, h = 0.1. y′ = x - y 2. ∴ f(x, y) = x - y 2. Modified Euler method. ym + 1 = ym + hf(xm + 1 2h, ym + 1 2hf(xm, ym)) f(x0, y0) = f(0, 1) = - 0.5. x0 + 1 2h = 0 + 0.1 2 = 0.05. y0 + 1 2hf(x0, y0) = 1 + 0.1 2 ⋅ - 0.5 = 0.975. f(x0 + 1 2h, y0 + 1 2hf(x0, y0) = f(0.05, 0.975) = - 0.4625.