how can i get an improved Euler's method code for this function?

15 次查看（过去 30 天）
Ibrahem abdelghany ghorab
评论： Santiago Cerón 2020-11-12
dy = @(x,y).2*x*y;
f = @(x).2*exp(x^2/2);
x0=1;
xn=1.5;
y=1;
h=0.1;
fprintf ('x \t \t y (euler)\t y(analytical) \n') % data table header
fprintf ('%f \t %f\t %f\n' ,x0,y,f(x0));
for x = x0 : h: xn-h
y = y + dy(x,y)*h;
x = x + h ;
fprintf (
'%f \t %f\t %f\n' ,x,y,f(x));
end
  2 个评论
FastCar
FastCar 2018-12-16
Euler has its limit to solve differential equations. You can change the integration step going towards the optimum step that is given by the minimum of the sum of the truncation error and step error, but you cannot improve further. What do you mean by improve?

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Are Mjaavatten
Are Mjaavatten 2018-12-17
There are two problems with your code:
  • The analytical solution is incorrect
  • You increment x inside the for loop. Don't. The for loop does this automatically.
Here is a corrected version:
a = 0.2;
y0 = 1;
x0 = 1;
xn = 1.5;
h = 0.1;
dy = @(x,y)a*x*y; % dy/dx
f = @(x) y0*exp(a/2*(x.^2-1)); % Correct analytic solution
y = y0;
fprintf ('x \t \t y (euler)\t y(analytical) \n') % data table header
fprintf ('%f \t %f\t %f\n' ,x0,y,f(x0));
for x = x0+h : h: xn
y = y + dy(x,y)*h;
fprintf ('%f \t %f\t %f\n' ,x,y,f(x));
end
Choose a smaller step length h to for better accuracy. Alternatively try a higher order method like Runge-Kutta.

更多回答（1 个）

James Tursa
James Tursa 2018-12-17
编辑：James Tursa 2018-12-17
The "Modified" Euler's Method is usually referring to the 2nd order scheme where you average the current and next step derivative in order to predict the next point. E.g.,
dy1 = dy(x,y); % derivative at this time point
dy2 = dy(x+h,y+h*dy1); % derivative at next time point from the normal Euler prediction
y = y + h * (dy1 + dy2) / 2; % average the two derivatives for the Modified Euler step
See this link:
https://en.wikipedia.org/wiki/Heun%27s_method
  4 个评论
显示 2更早的评论

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Get Started with MATLAB 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by