1 % % time step https://ww2.mathworks.cn/matlabcentral/answers/184200-newton-raphson-loop-for-backward-euler 2 % h = (t_final - t_init)/n; % with n number of time steps 3 % % vectors 4 % t = [tinit zeros(1,n)]; % time 5 % y = [yinit zeros(1,n)]; % solution 6 % % Backward Euler loop 7 % for i = 1:n 8 % t(i+1) = t(i) + h; 9 % y_temp = y(i) + h(f(t(i), y(i))); 10 % y(i+1) = y(i) + h*f(t(i+1), y_temp); 11 % end 12 % for i = 1:n 13 % error = 1; 14 % tolerance = 1e-6; 15 % t(i+1) = t(i) + h; 16 % y_temp = y(i) + h*(f(t(i), y(i))); 17 % while error >= tolerance 18 % y(i+1) = y(i) + h*f(t(i+1), y_temp); 19 % error = abs(y(i+1) - y_temp) % (local) absolute error 20 % y_temp = y(i+1); 21 % end 22 % end 23 24 % yold = y(i)+h*f(t(i),y(i)); 25 % while error >= tolerance 26 % ynew = yold-(yold-(y(i)+h*f(t(i+1),yold)))/(1-h*df(t(i+1),yold)); 27 % error = abs(ynew-yold); 28 % yold=ynew; 29 % end 30 % y(i+1) = ynew; 31 32 %y'=y+2*x/y^2 x=[0,2] y(0)=1 https://wenku.baidu.com/view/d18cdaa10b4c2e3f5627632f.html 33 t_final=2; 34 t_init=0; 35 n=5; 36 tolerance=0.0000001 37 h = (t_final - t_init)/n; 38 ti=t_init+h; 39 yold=1+h*f(0,1);% yold = y(i)+h*f(t(i),y(i)); 40 while error >= tolerance 41 ynew = yold-(yold-(y(i)+h*f(t(i+1),yold)))/(1-h*df(t(i+1),yold)); 42 error = abs(ynew-yold); 43 yold=ynew; 44 end 45 y(i+1) = ynew;
上面代码应该怎样修改？