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Home/ Questions/Q 823191
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T02:54:50+00:00 2026-05-15T02:54:50+00:00

Here is the MATLAB / FreeMat code I got to solve an ODE numerically

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Here is the MATLAB/FreeMat code I got to solve an ODE numerically using the backward Euler method. However, the results are inconsistent with my textbook results, and sometimes even ridiculously inconsistent. What is wrong with the code?

function [x,y] = backEuler(f,xinit,yinit,xfinal,h)

    %f - this is your y prime
    %xinit - initial X
    %yinit - initial Y
    %xfinal - final X
    %h - step size

    n = (xfinal-xinit)/h; %Calculate steps

    %Inititialize arrays...
    %The first elements take xinit and yinit corespondigly, the rest fill with 0s.
    x = [xinit zeros(1,n)];
    y = [yinit zeros(1,n)];

    %Numeric routine
    for i = 1:n
        x(i+1) = x(i)+h;
        ynew = y(i)+h*(f(x(i),y(i)));
        y(i+1) = y(i)+h*f(x(i+1),ynew);
    end
end
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  1. Editorial Team
    Editorial Team
    2026-05-15T02:54:50+00:00Added an answer on May 15, 2026 at 2:54 am

    Your method is a method of a new kind. It is neither backward nor forward Euler. 🙂

    Forward Euler: y1 = y0 + h*f(x0,y0)

    Backward Euler solve in y1: y1 - h*f(x1,y1) = y0

    Your method: y1 = y0 +h*f(x0,x0+h*f(x0,y0))

    Your method is not backward Euler.

    You don’t solve in y1, you just estimate y1 with the forward Euler method. I don’t want to pursue the analysis of your method, but I believe it will behave poorly indeed, even compared with forward Euler, since you evaluate the function f at the wrong point.

    Here is the closest method to your method that I can think of, explicit as well, which should give much better results. It’s Heun’s Method:

    y1 = y0 + h/2*(f(x0,y0) + f(x1,x0+h*f(x0,y0)))

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