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

23 vues (au cours des 30 derniers jours)
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 commentaires
FastCar
FastCar le 16 Déc 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?
Ibrahem abdelghany ghorab
Ibrahem abdelghany ghorab le 17 Déc 2018
modified method
and i what 4Runge-kutta for this function dy = @(x,y).2*x*y;

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Réponse acceptée

Are Mjaavatten
Are Mjaavatten le 17 Déc 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 commentaire
Ibrahem abdelghany ghorab
Ibrahem abdelghany ghorab le 17 Déc 2018
Modifié(e) : Ibrahem abdelghany ghorab le 17 Déc 2018
modified orImprovedEuler method
and i what 4Runge-kutta for this function dy = @(x,y).2*x*y;

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Plus de réponses (1)

James Tursa
James Tursa le 17 Déc 2018
Modifié(e) : James Tursa le 17 Déc 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 commentaires
Ibrahem abdelghany ghorab
Ibrahem abdelghany ghorab le 18 Déc 2018
thank you very much
Santiago Cerón
Santiago Cerón le 12 Nov 2020
James, how do you graph that in a plot?

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