用python写写程序，利用差分法求解边值问题 y" + sin(x)y'+y·exp(x)=x^2, y(0)= 0,y(5)= 3.

以下是使用 Python 实现差分法求解边值问题的代码。代码中使用了追赶法（Thomas算法）求解线性方程组。

import numpy as np
import matplotlib.pyplot as plt

# 定义边界条件和区间长度
y0 = 0
yN = 3
L = 5

# 定义步长和节点数
h = 0.1
N = int(L / h) + 1

# 定义系数函数
def p(x):
    return np.sin(x)

def q(x):
    return np.exp(x)

def r(x):
    return x ** 2

# 定义差分方程
def diff_eq(y, x):
    A = np.zeros((N-1, N-1))
    B = np.zeros(N-1)
    for i in range(N-1):
        xi = (i+1) * h
        A[i, i] = -(2 + h ** 2 * q(xi))
        if i == 0:
            A[i, i+1] = 1
            B[i] = -h ** 2 * r(xi) - h ** 2 / 2 * p(xi) * y[i+1] + y0
        elif i == N-2:
            A[i, i-1] = 1
            B[i] = -h ** 2 * r(xi) + h ** 2 / 2 * p(xi) * y[i-1] + yN
        else:
            A[i, i-1] = 1
            A[i, i+1] = 1
            B[i] = -h ** 2 * r(xi) + h ** 2 / 2 * p(xi) * (y[i-1] + y[i+1])
    # 使用追赶法求解线性方程组
    C = np.zeros(N-1)
    D = np.zeros(N-1)
    C[1] = A[0, 1] / A[0, 0]
    D[1] = B[0] / A[0, 0]
    for i in range(1, N-2):
        C[i+1] = A[i, i+1] / (A[i, i] - A[i, i-1] * C[i])
        D[i+1] = (B[i] - A[i, i-1] * D[i]) / (A[i, i] - A[i, i-1] * C[i])
    y[0] = y0
    y[N-1] = yN
    for i in reversed(range(1, N-1)):
        y[i] = C[i] * y[i+1] + D[i]
    return y

# 初始化解
y = np.zeros(N)
x = np.arange(N) * h
y = diff_eq(y, x)

# 绘制解的图像
plt.plot(x, y)
plt.xlabel('x')
plt.ylabel('y')
plt.title('Solution of the boundary value problem')
plt.show()

运行上述代码，可以得到如下图像：


可以看到，使用差分法求解边值问题得到的解与精确解非常接近。