how can i get an improved Euler's method code for this function?
22 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Ibrahem abdelghany ghorab
el 15 de Dic. de 2018
Comentada: Santiago Cerón
el 12 de Nov. de 2020
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 comentarios
FastCar
el 16 de Dic. de 2018
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?
Respuesta aceptada
Are Mjaavatten
el 17 de Dic. de 2018
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.
1 comentario
Ibrahem abdelghany ghorab
el 17 de Dic. de 2018
Editada: Ibrahem abdelghany ghorab
el 17 de Dic. de 2018
Más respuestas (1)
James Tursa
el 17 de Dic. de 2018
Editada: James Tursa
el 17 de Dic. de 2018
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:
4 comentarios
Ver también
Categorías
Más información sobre Calculus en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!