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

 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  

[FastCar]

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;

Answer 1

[Are Mjaavatten]

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.

[Ibrahem abdelghany ghorab]

modified orImprovedEuler method

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

Answer 2

[James Tursa]

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

[Ibrahem abdelghany ghorab]

can you write the all code please

[James Tursa]

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.

[Ibrahem abdelghany ghorab]

thank you very much

[Santiago Cerón]

James, how do you graph that in a plot?