数值分析 - 编程作业(6)

N=length(yadms); p=zeros(N,1); m=p; c=m; for i=4:n

p(i+1)=yadms(i)+(h/24)*(55*fxy(x(i),yadms(i))-59*fxy(x(i-1),...

yadms(i-1))+37*fxy(x(i-2),yadms(i-2))-9*fxy(x(i-3),yadms(i-3))); m(i+1)=p(i+1)+(251/270)*(c(i)-p(i));

c(i+1)=yadms(i)+(h/24)*(9*fxy(x(i+1),m(i+1))+19*fxy(x(i),...

yadms(i))-5*fxy(x(i-1),yadms(i-1))+fxy(x(i-2),yadms(i-2))); yadms(i+1)=c(i+1)-(19/270)*(c(i+1)-p(i+1)); end

%子函数---fxy函数 function f=fxy(x,y) f=-y+2*cos(x);

%主程序---Adams预测校正算法 clear clc a=0; b=pi; N=50; y0=1;

yrk4=rk4(a,b,y0,N,4);

yadms=adms(a,b,yrk4(1:4),N); %绘图---精确解

y=dsolve('Dy=-y+2*cos(x)','y(0)=1','x'); x1=0:(pi/N):pi; y1=subs(y,x1); plot(x1,y1,'-b') hold on

%绘图---数值解 plot(x1,yadms,'xr') grid on

legend('精确解','数值解')

title(Adams预测校正数值解与精确解')

运行结果：

Adams预测校正数值解与精确解1.5精确解数值解1 0.50-0.5-1-1.5 00.511.522.533.5由实验结果可以看出，该2种算法是稳定的。求解效果良好。