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时间:2019-08-25
《方程的欧拉法》由会员上传分享,免费在线阅读,更多相关内容在工程资料-天天文库。
1、2013-2014(1)专业课程实践论文题目:方程的欧拉法一、算法理论我们知道,在0小平而上,微分方程ya2、折线的极点巳,过人(£,儿)依方向场的方向再推进到垃+[(£+“儿+J,显然两个极点出,為的坐标有以下关系:儿》_儿=/(£,儿)陥一兀即儿+】=儿+僻(£,儿)。这就是著名的Euler格式。若初始值y()己知,则格式斤=0丄2,…可逐步算出二.算法框图三.算法程序#include"stdio.h"#include,^math.h'^#include"conio.h"#include,,stdlib.hH#defineN11doublef(doublex,doubley){returny・2*x/y;}voidEule3、r(){doublex[N],y[N],z[N],e[N];intn;system("clsH);printf(”『);for(n=0;n4、n]=%5.4fz[n]=%5.4fe[n]=%5.4fM5、,y[n],z6、n],e[n]);}getch();return;}voidmain(){intc;while(1){system(Hcls");puts(u1.Euler:n);puts("0.exitAn*');puts(”choice:");c=getchar();switch(c){caseT:Euler();break;caseO:system(HclsH);exit(O);default:break;四、算法实现例1•使用公式,求一阶常微分方程初值问题2x小.y二y-07、h=OAo解:(1)输入初始值1(2)显示结果wE:MicrosoftVisualStudioMyProjectssdfDebugsdf.exe"[nJ=1.0000&[nl=1.1000&[n]=1.1918&[nJ=1.2774&[n)=1.3582p[nl=1.4351&[n)=1.5090&[nl=l.5803&[nl=1.6498&[n1=1.7178&[nl=1.7848z[n1=1.0000ztn1=1.0954z[nl=1.1832ztn3=1.2649z[nJ=1.3416ztn1=1.418、42ztn]=1.4832ztn3=1.5492z[n]=1.6125ztn3=1.6733ztn]=1.7321etn]=0.0000e[nJ=0.0046e[nJ=0.0086eCn1=0.0125etnJ=0.0166e[nJ=0.0209e[nJ=0.0257eCn1=0.0311etn]=0.0373etn]=0.0445etn]=0.0527例2•使用EMe公式,求一阶常微分方程初值问题y=y-2厂(0<%<1)y(0)=1的数值解,力二0.1。解:(1)输入初始值1(2)显不结果E:MicrosoftV9、isualStudioMyProjectssdfDebugsdf.exe"ytnJsl.0000y[nl=1.1000y[nl=1.1880y[nJ=l.2593ytn3=1.3097SFCnJ=1.3358y[nW3358y[nl=1.3091y[n]=l.2568y[nJ=1.1814ytn1^1.0868z[n3=1.0000z[nl=1.0954z[nl=1.1832zln)=1.2649z[nl=1.3416z[n]=1.4142ztn1=1.4832zfn1=1.5492z[nJ=1.6125z[n10、l=1.6?33zEn1=1.7321eLn]=0.0000eCn]=0.0046e[n]=0.0048etn]=-0.0056eEn]=-0.0320etn]=-0.0784etn]=-0.1474eCn]=-0.2401eEn]=-0.3557eEn]=-0.4920e5]=—0・6452
2、折线的极点巳,过人(£,儿)依方向场的方向再推进到垃+[(£+“儿+J,显然两个极点出,為的坐标有以下关系:儿》_儿=/(£,儿)陥一兀即儿+】=儿+僻(£,儿)。这就是著名的Euler格式。若初始值y()己知,则格式斤=0丄2,…可逐步算出二.算法框图三.算法程序#include"stdio.h"#include,^math.h'^#include"conio.h"#include,,stdlib.hH#defineN11doublef(doublex,doubley){returny・2*x/y;}voidEule
3、r(){doublex[N],y[N],z[N],e[N];intn;system("clsH);printf(”『);for(n=0;n4、n]=%5.4fz[n]=%5.4fe[n]=%5.4fM5、,y[n],z6、n],e[n]);}getch();return;}voidmain(){intc;while(1){system(Hcls");puts(u1.Euler:n);puts("0.exitAn*');puts(”choice:");c=getchar();switch(c){caseT:Euler();break;caseO:system(HclsH);exit(O);default:break;四、算法实现例1•使用公式,求一阶常微分方程初值问题2x小.y二y-07、h=OAo解:(1)输入初始值1(2)显示结果wE:MicrosoftVisualStudioMyProjectssdfDebugsdf.exe"[nJ=1.0000&[nl=1.1000&[n]=1.1918&[nJ=1.2774&[n)=1.3582p[nl=1.4351&[n)=1.5090&[nl=l.5803&[nl=1.6498&[n1=1.7178&[nl=1.7848z[n1=1.0000ztn1=1.0954z[nl=1.1832ztn3=1.2649z[nJ=1.3416ztn1=1.418、42ztn]=1.4832ztn3=1.5492z[n]=1.6125ztn3=1.6733ztn]=1.7321etn]=0.0000e[nJ=0.0046e[nJ=0.0086eCn1=0.0125etnJ=0.0166e[nJ=0.0209e[nJ=0.0257eCn1=0.0311etn]=0.0373etn]=0.0445etn]=0.0527例2•使用EMe公式,求一阶常微分方程初值问题y=y-2厂(0<%<1)y(0)=1的数值解,力二0.1。解:(1)输入初始值1(2)显不结果E:MicrosoftV9、isualStudioMyProjectssdfDebugsdf.exe"ytnJsl.0000y[nl=1.1000y[nl=1.1880y[nJ=l.2593ytn3=1.3097SFCnJ=1.3358y[nW3358y[nl=1.3091y[n]=l.2568y[nJ=1.1814ytn1^1.0868z[n3=1.0000z[nl=1.0954z[nl=1.1832zln)=1.2649z[nl=1.3416z[n]=1.4142ztn1=1.4832zfn1=1.5492z[nJ=1.6125z[n10、l=1.6?33zEn1=1.7321eLn]=0.0000eCn]=0.0046e[n]=0.0048etn]=-0.0056eEn]=-0.0320etn]=-0.0784etn]=-0.1474eCn]=-0.2401eEn]=-0.3557eEn]=-0.4920e5]=—0・6452
4、n]=%5.4fz[n]=%5.4fe[n]=%5.4fM
5、,y[n],z
6、n],e[n]);}getch();return;}voidmain(){intc;while(1){system(Hcls");puts(u1.Euler:n);puts("0.exitAn*');puts(”choice:");c=getchar();switch(c){caseT:Euler();break;caseO:system(HclsH);exit(O);default:break;四、算法实现例1•使用公式,求一阶常微分方程初值问题2x小.y二y-07、h=OAo解:(1)输入初始值1(2)显示结果wE:MicrosoftVisualStudioMyProjectssdfDebugsdf.exe"[nJ=1.0000&[nl=1.1000&[n]=1.1918&[nJ=1.2774&[n)=1.3582p[nl=1.4351&[n)=1.5090&[nl=l.5803&[nl=1.6498&[n1=1.7178&[nl=1.7848z[n1=1.0000ztn1=1.0954z[nl=1.1832ztn3=1.2649z[nJ=1.3416ztn1=1.418、42ztn]=1.4832ztn3=1.5492z[n]=1.6125ztn3=1.6733ztn]=1.7321etn]=0.0000e[nJ=0.0046e[nJ=0.0086eCn1=0.0125etnJ=0.0166e[nJ=0.0209e[nJ=0.0257eCn1=0.0311etn]=0.0373etn]=0.0445etn]=0.0527例2•使用EMe公式,求一阶常微分方程初值问题y=y-2厂(0<%<1)y(0)=1的数值解,力二0.1。解:(1)输入初始值1(2)显不结果E:MicrosoftV9、isualStudioMyProjectssdfDebugsdf.exe"ytnJsl.0000y[nl=1.1000y[nl=1.1880y[nJ=l.2593ytn3=1.3097SFCnJ=1.3358y[nW3358y[nl=1.3091y[n]=l.2568y[nJ=1.1814ytn1^1.0868z[n3=1.0000z[nl=1.0954z[nl=1.1832zln)=1.2649z[nl=1.3416z[n]=1.4142ztn1=1.4832zfn1=1.5492z[nJ=1.6125z[n10、l=1.6?33zEn1=1.7321eLn]=0.0000eCn]=0.0046e[n]=0.0048etn]=-0.0056eEn]=-0.0320etn]=-0.0784etn]=-0.1474eCn]=-0.2401eEn]=-0.3557eEn]=-0.4920e5]=—0・6452
7、h=OAo解:(1)输入初始值1(2)显示结果wE:MicrosoftVisualStudioMyProjectssdfDebugsdf.exe"[nJ=1.0000&[nl=1.1000&[n]=1.1918&[nJ=1.2774&[n)=1.3582p[nl=1.4351&[n)=1.5090&[nl=l.5803&[nl=1.6498&[n1=1.7178&[nl=1.7848z[n1=1.0000ztn1=1.0954z[nl=1.1832ztn3=1.2649z[nJ=1.3416ztn1=1.41
8、42ztn]=1.4832ztn3=1.5492z[n]=1.6125ztn3=1.6733ztn]=1.7321etn]=0.0000e[nJ=0.0046e[nJ=0.0086eCn1=0.0125etnJ=0.0166e[nJ=0.0209e[nJ=0.0257eCn1=0.0311etn]=0.0373etn]=0.0445etn]=0.0527例2•使用EMe公式,求一阶常微分方程初值问题y=y-2厂(0<%<1)y(0)=1的数值解,力二0.1。解:(1)输入初始值1(2)显不结果E:MicrosoftV
9、isualStudioMyProjectssdfDebugsdf.exe"ytnJsl.0000y[nl=1.1000y[nl=1.1880y[nJ=l.2593ytn3=1.3097SFCnJ=1.3358y[nW3358y[nl=1.3091y[n]=l.2568y[nJ=1.1814ytn1^1.0868z[n3=1.0000z[nl=1.0954z[nl=1.1832zln)=1.2649z[nl=1.3416z[n]=1.4142ztn1=1.4832zfn1=1.5492z[nJ=1.6125z[n
10、l=1.6?33zEn1=1.7321eLn]=0.0000eCn]=0.0046e[n]=0.0048etn]=-0.0056eEn]=-0.0320etn]=-0.0784etn]=-0.1474eCn]=-0.2401eEn]=-0.3557eEn]=-0.4920e5]=—0・6452
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