1 #include<stdio.h>
 2 #include<math.h>
 3 #define n 14
 4 //double func1(double x, double y);
 5 double func2(double x, double y);
 6 int main(){
 7     double a = 0.0, b = 1.4;//求解区间为[a,b]
 8     double h, m = 0.0;//h为步长，取0.1；  m为y的初值。
 9     double x[n + 1] = { 0.0 }, y[n + 1] = { 0.0 };//初始化数组
10     h = (b - a) / n;//求步长
11     x[0] = a;
12     y[0] = m;
13     printf("要求解的常微分方程为：dy/dx=1+y^2,初值为y(0)=0.0;\n");
14     printf("下面为预估校正法结果：\n");
15     //下面为预估校正法（即向前欧拉法和梯形法的结合，也称改进的欧拉法）
16     for (int i = 1; i <= n; i++)
17     {
18         y[i] = y[i - 1] + h*(1 + y[i - 1] * y[i - 1]);
19         x[i] = a + i*h;
20         y[i] = y[i - 1] + h*0.5*((1 + y[i - 1] * y[i - 1]) + (1 + y[i] * y[i]));
21         printf("x[%d]=%e , y[%d]=%e\n", i, x[i], i, y[i]);
22     }
23     printf("下面为四阶龙格-库塔法结果：\n");
24     double k1 = 0.0, k2 = 0.0, k3 = 0.0, k4 = 0.0;
25     for (int i = 1; i <= n; i++)
26     {
27         x[i] = 0.0, y[i] = 0.0;
28     }
29     x[0] = a;
30     y[0] = m;
31     for (int i = 1; i <= n; i++)
32     {
33         k1 = h*func2(x[i - 1], y[i - 1]);
34         k2 = h*func2(x[i - 1] + h / 2.0, y[i - 1] + 0.5*k1);
35         k3 = h*func2(x[i - 1] + h / 2.0, y[i - 1] + 0.5*k2);
36         k4 = h*func2(x[i - 1] + h, y[i - 1] + k3);
37         y[i] = y[i - 1] + (1.0 / 6.0)*(k1 + 2 * k2 + 2 * k3 + k4);
38         printf("x[%d]=%e , y[%d]=%e\n", i, x[i], i, y[i]);
39         x[i] = a + i*h;
40     }
41     return 0;
42 }
43 /*
44 double func1(double x, double y)//dy/dx=x-y+1,func1=x-y+1;
45 {
46 double f = 0;
47 f = x - y + 1;
48 return f;
49 }
50 */
51 double func2(double x, double y)//dy/dx=1+y*y,func2=1+y*y;
52 {
53     double f = 0.0;
54     f = 1 + y*y;
55     return f;
56 }