3次初期値問題の常微分方程式を解く

import numpy as np
from matplotlib import pyplot as plt
plt.rcParams['font.sans-serif']=['SimHei']
plt.rcParams['axes.unicode_minus'] = False


def myfun1(x, y1, y2, y3):
    return y2

def myfun2(x, y1, y2, y3):
    return y3

def myfun3(x,y1, y2, y3):
    return 3*x*x


def ode(myfun1, myfun2, myfun3, x, y1, y2, y3, h, b):
    y1_r = []
    y2_r = []
    y3_r = []
    x_r = []
    y1_r.append(y1)
    y2_r.append(y2)
    y3_r.append(y3)
    x_r.append(x)
    for i in range(int(b / h)):
        k11 = h * myfun1(x, y1, y2, y3)
        k12 = h * myfun2(x, y1, y2, y3)
        k13 = h * myfun3(x, y1, y2, y3)
        k21 = h * myfun1(x + h / 2, y1 + 0.5 * k11, y2 + 0.5 * k12, y3 + 0.5 * k13)
        k22 = h * myfun2(x + h / 2, y1 + 0.5 * k11, y2 + 0.5 * k12, y3 + 0.5 * k13)
        k23 = h * myfun3(x + h / 2, y1 + 0.5 * k11, y2 + 0.5 * k12, y3 + 0.5 * k13)
        k31 = h * myfun1(x + h / 2, y1 + 0.5 * k21, y2 + 0.5 * k22, y3 + 0.5 * k23)
        k32 = h * myfun2(x + h / 2, y1 + 0.5 * k21, y2 + 0.5 * k22, y3 + 0.5 * k23)
        k33 = h * myfun3(x + h / 2, y1 + 0.5 * k21, y2 + 0.5 * k22, y3 + 0.5 * k23)
        k41 = h * myfun1(x + h, y1 + k31, y2 + k32, y3 + k33)
        k42 = h * myfun2(x + h, y1 + k31, y2 + k32, y3 + k33)
        k43 = h * myfun3(x + h, y1 + k31, y2 + k32, y3 + k33)
        y1 = y1 + (k11 + 2 * k21 + 2 * k31 + k41) / 6
        y2 = y2 + (k12 + 2 * k22 + 2 * k32 + k42) / 6
        y3 = y3 + (k13 + 2 * k23 + 2 * k33 + k43) / 6
        y1_r.append(y1)
        y2_r.append(y2)
        y3_r.append(y3)
        x = x + h
        x_r.append(x)
    return x_r, y1_r, y2_r,y3_r

x = np.linspace(0, 1, 20)
y = x**4/4
x_r, y1_r, y2_r, y3_r= ode(myfun1, myfun2,myfun3, 0, 0, 0, 0, 0.001, 1)
plt.plot(x, y, 'ks:', markersize=3)
plt.plot(x_r, y1_r, 'b*:', markersize=1.5)
plt.plot(x_r, y2_r, 'g^:', markersize=1.5)
plt.plot(x_r, y3_r, 'r>:', markersize=1.5)
plt.legend(['函数值数值计算结果', '原函数', '导数数值计算结果','二阶导数数值计算结果'])
plt.show()