数值计算上机第七题关于改进欧拉方法
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infile>>x0>>x1>>h>>y0;//依次输入x的初值x0，x的终值x1，步长h，y（0）的值y0
outfile<<x0<<" "<<y0<<endl;
while(x0<x1-h)
{
yp=y0+h*sqrt(2*x0*x0+3*y0*y0);
x0=x0+h;
yc=y0+h*sqrt(2*x0*x0+3*yp*yp);
13.4 4.19738e+010
13.6 5.90324e+010
13.8 8.30238e+010
14 1.16766e+011
14.2 1.6422e+011
14.4 2.30961e+011
14.6 3.24826e+011
14.8 4.56839e+011
15 6.42502e+011
15.2 9.03622e+011
15.4 1.27086e+012
15.6 1.78735e+012
15.8 2.51375e+012
16 3.53537e+012
16.2 4.97218e+012
16.4 6.99292e+012
16.6 9.83492e+012
16.8 1.38319e+013
17 1.94534e+013
17.2 2.73594e+013
1.4 54.4735
1.6 76.6169
1.8 107.759
2 151.558
2.2 213.156
2.4 299.787
2.6 421.626
2.8 592.981
3 833.977
3.2 1172.91
3.4 1649.6
3.6 2320.01
3.8 3262.89
4 4588.97
4.2 6453.97
y0=1.0/2*(yp+yc);
outfile<<x0<<" "<<y0<<endl;
}//改进欧拉法公式并计算结果
return 0;
}
输入:
0 20 0.2 5
输出:
0 5
0.2 7.03239
0.4 9.89185
0.6 13.9148
0.8 19.5739
1 27.5337
1.2 38.7287
7.取h=0.2，用改进欧拉方法求解下列初值问题。
( )
第七题:
#include "fstream.h"
#include "math.h"
int main()
{
double x0,x1,y0,yp,yc,h;
ifstream infile("in.dat");
ofstream outfile("out.dat");
19.4 1.16503e+015
19.6 1.63852e+015
19.8 2.30443e+015
20 3.24097e+015
17.4 3.84785e+013
17.6 5.41166e+013
17.8 7.61102e+013
18 1.07042e+014
18.2 1.50545e+014
18.4 2.11728e+014
18.6 2.97777e+014
18.8 4.18796e+014
19 5.88999e+014
19.2 8.28374e+014
11.4 1.38631e+009
11.6 1.94971e+009
11.8 2.7421e+009
12 3.85651e+009
12.2 5.42384e+009
12.4 7.62814e+009
12.6 1.07283e+010
12.8 1.50884e+010
13 2.12205e+010
13.2 2.98447e+010
9.4 4.57866e源自文库007
9.6 6.43948e+007
9.8 9.05655e+007
10 1.27372e+008
10.2 1.79138e+008
10.4 2.51941e+008
10.6 3.54332e+008
10.8 4.98337e+008
11 7.00866e+008
11.2 9.85705e+008
4.4 9076.93
4.6 12765.9
4.8 17954.1
5 25250.8
5.2 35513
5.4 49945.8
5.6 70244.3
5.8 98792.3
6 138942
6.2 195410
6.4 274827
6.6 386519
6.8 543604
7 764531
7.2 1.07524e+006
7.4 1.51223e+006
7.6 2.12682e+006
7.8 2.99118e+006
8 4.20683e+006
8.2 5.91653e+006
8.4 8.32106e+006
8.6 1.17028e+007
8.8 1.6459e+007
9 2.31481e+007
9.2 3.25557e+007
outfile<<x0<<" "<<y0<<endl;
while(x0<x1-h)
{
yp=y0+h*sqrt(2*x0*x0+3*y0*y0);
x0=x0+h;
yc=y0+h*sqrt(2*x0*x0+3*yp*yp);
13.4 4.19738e+010
13.6 5.90324e+010
13.8 8.30238e+010
14 1.16766e+011
14.2 1.6422e+011
14.4 2.30961e+011
14.6 3.24826e+011
14.8 4.56839e+011
15 6.42502e+011
15.2 9.03622e+011
15.4 1.27086e+012
15.6 1.78735e+012
15.8 2.51375e+012
16 3.53537e+012
16.2 4.97218e+012
16.4 6.99292e+012
16.6 9.83492e+012
16.8 1.38319e+013
17 1.94534e+013
17.2 2.73594e+013
1.4 54.4735
1.6 76.6169
1.8 107.759
2 151.558
2.2 213.156
2.4 299.787
2.6 421.626
2.8 592.981
3 833.977
3.2 1172.91
3.4 1649.6
3.6 2320.01
3.8 3262.89
4 4588.97
4.2 6453.97
y0=1.0/2*(yp+yc);
outfile<<x0<<" "<<y0<<endl;
}//改进欧拉法公式并计算结果
return 0;
}
输入:
0 20 0.2 5
输出:
0 5
0.2 7.03239
0.4 9.89185
0.6 13.9148
0.8 19.5739
1 27.5337
1.2 38.7287
7.取h=0.2，用改进欧拉方法求解下列初值问题。
( )
第七题:
#include "fstream.h"
#include "math.h"
int main()
{
double x0,x1,y0,yp,yc,h;
ifstream infile("in.dat");
ofstream outfile("out.dat");
19.4 1.16503e+015
19.6 1.63852e+015
19.8 2.30443e+015
20 3.24097e+015
17.4 3.84785e+013
17.6 5.41166e+013
17.8 7.61102e+013
18 1.07042e+014
18.2 1.50545e+014
18.4 2.11728e+014
18.6 2.97777e+014
18.8 4.18796e+014
19 5.88999e+014
19.2 8.28374e+014
11.4 1.38631e+009
11.6 1.94971e+009
11.8 2.7421e+009
12 3.85651e+009
12.2 5.42384e+009
12.4 7.62814e+009
12.6 1.07283e+010
12.8 1.50884e+010
13 2.12205e+010
13.2 2.98447e+010
9.4 4.57866e源自文库007
9.6 6.43948e+007
9.8 9.05655e+007
10 1.27372e+008
10.2 1.79138e+008
10.4 2.51941e+008
10.6 3.54332e+008
10.8 4.98337e+008
11 7.00866e+008
11.2 9.85705e+008
4.4 9076.93
4.6 12765.9
4.8 17954.1
5 25250.8
5.2 35513
5.4 49945.8
5.6 70244.3
5.8 98792.3
6 138942
6.2 195410
6.4 274827
6.6 386519
6.8 543604
7 764531
7.2 1.07524e+006
7.4 1.51223e+006
7.6 2.12682e+006
7.8 2.99118e+006
8 4.20683e+006
8.2 5.91653e+006
8.4 8.32106e+006
8.6 1.17028e+007
8.8 1.6459e+007
9 2.31481e+007
9.2 3.25557e+007