№8 Метод Эйлера решения задачи Коши для обыкновенного дифференциального уравнения (ОДУ) 1-го порядка

import math

eps = 1e-6
V = 8


def f(x, y):
    return 3 * x * x - 2 * V * x


def check_y(x):
    return x*x*(x - V)


def euler(n, h, x0, y0):
    x = [0] * n
    x[0] = x0
    for i in range(n - 1):
        x[i + 1] = x[i] + h
    y = [0] * n
    y[0] = y0
    for i in range(n - 1):
        y[i + 1] = y[i] + h * f(x[i], y[i])
    return y


def ex_euler(n, h, x0, y0):
    x = [0] * (2 * n)
    x[0] = x0
    for i in range(2 * n - 1):
        x[i + 1] = x[i] + h / 2
    y = [0] * (2 * n)
    y[0] = y0
    for i in range(2 * n - 1):
        if i % 2:
            y[i + 1] = y[i - 1] + h * f(x[i], y[i])
        else:
            y[i + 1] = y[i] + h / 2 * f(x[i], y[i])
    res = [0] * n
    for i in range(n):
        res[i] = y[2 * i]
    return res


def print_result(vx, vy):
    for x in vx:
        print(f"{x:.2f}", end="\t")
    print()
    for x in vy:
        print(f"{x:.2f}", end="\t")
    print()
    for x in vx:
        print(f"{check_y(x):.2f}", end="\t")
    print()
    for i in range(len(vy)):
        print(f"{vy[i] - check_y(vx[i]):.2f}", end="\t")
    print()
    mx = 0
    for i in range(len(vy)):
        mx = max(mx, abs(vy[i] - check_y(vx[i])))
    print(f"Максимальное отклонение: {mx:.5f}")


def solve():
    n = 13
    h = 1e-0
    x0 = 1
    y0 = 0
    vx = [0] * n
    vx[0] = x0
    for i in range(1, n):
        vx[i] = vx[i - 1] + h

    print("Метод Эйлера:")
    vy1 = euler(n, h, x0, y0)
    print_result(vx, vy1)
    print()

    print("Расширенный метод Эйлера:")
    vy2 = ex_euler(n, h, x0, y0)
    print_result(vx, vy2)
    print()


if __name__ == "__main__":
    solve()