% % time step https://ww2.mathworks.cn/matlabcentral/answers/184200-newton-raphson-loop-for-backward-euler
% h = (t_final - t_init)/n; % with n number of time steps
% % vectors
% t = [tinit zeros(,n)]; % time
% y = [yinit zeros(,n)]; % solution
% % Backward Euler loop
% for i = :n
% t(i+) = t(i) + h;
% y_temp = y(i) + h(f(t(i), y(i)));
% y(i+) = y(i) + h*f(t(i+), y_temp);
% end
% for i = :n
% error = ;
% tolerance = 1e-;
% t(i+) = t(i) + h;
% y_temp = y(i) + h*(f(t(i), y(i)));
% while error >= tolerance
% y(i+) = y(i) + h*f(t(i+), y_temp);
% error = abs(y(i+) - y_temp) % (local) absolute error
% y_temp = y(i+);
% end
% end % yold = y(i)+h*f(t(i),y(i));
% while error >= tolerance
% ynew = yold-(yold-(y(i)+h*f(t(i+),yold)))/(-h*df(t(i+),yold));
% error = abs(ynew-yold);
% yold=ynew;
% end
% y(i+) = ynew; %y'=y+2*x/y^2 x=[0,2] y(0)=1 https://wenku.baidu.com/view/d18cdaa10b4c2e3f5627632f.html
t_final=;
t_init=;
n=;
tolerance=0.0000001
h = (t_final - t_init)/n;
ti=t_init+h;
yold=+h*f(,);% yold = y(i)+h*f(t(i),y(i));
while error >= tolerance
ynew = yold-(yold-(y(i)+h*f(t(i+),yold)))/(-h*df(t(i+),yold));
error = abs(ynew-yold);
yold=ynew;
end
y(i+) = ynew;
上面代码应该怎样修改？