Why is the MATLAB output of this numerical method precision not getting more accurate?

Question

Posting here vs math.stackexchange because I think my issue is syntax:

I'm trying to analyze the 2nd order ODE: y'' + 2y' + 2y = e^(-x) * sin(x) using MATLAB code for the midpoint method. I first converted the ODE to a system of 1st order equations and then tried to apply it below, but as the discretizations [m] are increased, the output is stopping at .2718. For example, m=11 yields:

ans =

0.2724

and m=101:

ans =

0.2718

and m=10001

ans =

0.2718

Here's the code:

function [y,t] = ODEsolver_midpointND(F,y0,a,b,m)

    if nargin < 5, m = 11; end
    if nargin < 4, a = 0; b = 1; end
    if nargin < 3, a = 0; b = 1; end
    if nargin < 2, error('invalid number of inputs'); end

    t = linspace(a,b,m)';
    h = t(2)-t(1);

    n = length(y0);

    y = zeros(m,n);
    y(1,:) = y0;
    for i=2:m
    Fty = feval(F,t(i-1),y(i-1,:));
    th = t(i-1)+h/2;

    y(i,:) = y(i-1,:) + ...
        h*feval(F,th,y(i-1,:)+(h/2)*Fty );
    end

Separate file:

function F = Fexample1(t,y)

F1 = y(2);
F2 = exp(-t).*sin(t)-2.*y(2)-2.*y(1);
F = [F1,F2];

Third file:

[Y,t] = ODEsolver_midpointND('Fexample1',[0 0],0,1,11);
Ye = [(1./2).*exp(-t).*(sin(t)-t.*cos(t)) (1./2).*exp(-t).*((t-1).*sin(t)-  t.*cos(t))];
norm(Y-Ye,inf)

Answer 1

Your ODE solver looks to me like it should work - however there's a typo in the analytic solution you're comparing to. It should be

Ye = [(1./2).*exp(-t).*(sin(t)-t.*cos(t)) (1./2).*exp(-t).*((t-1).*sin(t)+  t.*cos(t))];

i.e. with a + sign before the t.*cos(t) term in the derivative.