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

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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

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FastCar
FastCar 16 Dec 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
modified method
and i what 4Runge-kutta for this function dy = @(x,y).2*x*y;

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채택된 답변

Are Mjaavatten
Are Mjaavatten 17 Dec 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.

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Ibrahem abdelghany ghorab
modified orImprovedEuler method
and i what 4Runge-kutta for this function dy = @(x,y).2*x*y;

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More Answers (1)

James Tursa
James Tursa 17 Dec 2018
James Tursa 님이 편집함. 17 Dec 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:

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James Tursa
James Tursa 18 Dec 2018
Not sure what you are asking. The loop is simply
for x = x0 : h: xn-h
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
fprintf ('%f \t %f\t %f\n' ,x+h,y,f(x+h));
end
Note that inside the fprintf I have used x+h, since that is the x value associated with the newly calculated y value.

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